March 14, 2017 | Author: ΘωμαςΣτεφανιδης | Category: N/A
Volume 23 Managing Editor Mahabir Singh Editor Anil Ahlawat (BE, MBA)
No. 11
November 2015
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CONTENTS
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Physics Musing Problem Set 28
8
JEE Workouts
12
Core Concept
18
Physics Musing Solution Set 27
21
Exam Prep 2016
23
JEE Accelerated Learning Series Brain Map
31 46
JEE Advanced Practice Paper 2016
58
Ace Your Way CBSE XI
64
Thought Provoking Problems
71
Ace Your Way CBSE XII
74
You Ask We Answer
82
Live Physics
83
Crossword
85
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Physics for you | November ‘15
7
P
PHYSICS
MUSING
hysics Musing was started in August 2013 issue of Physics For You with the suggestion of Shri Mahabir Singh. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / AIIMS / Other PMTs with additional study material. In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / various PMTs. The detailed solutions of these problems will be published in next issue of Physics For You. The readers who have solved five or more problems may send their detailed solutions with their names and complete address. The names of those who send atleast five correct solutions will be published in the next issue. We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams.
single oPtion correct tyPe
1. A metal ring of initial radius r and cross-sectional area A is fitted onto a wooden disc of radius R > r. If Young’s modulus of the metal is Y, then the tension in the ring is (a) AYR (b) Yr AR r (c)
AY (R − r ) r
(d)
Y (R − r ) Ar
2. A piece of pure gold (r = 19.3 g cm–3) is suspected to be hollow from inside. It weighs 38.250 g in air and 33.865 g in water. The volume of the hollow portion in gold is (a) 1.982 cm3 (b) 2.403 cm3 (c) 3.825 cm3 (d) 4.385 cm3 3. A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats g. It is moving with speed v and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by (a)
( g − 1) Mv 2 2( g + 2)R
(b)
( g −1) Mv 2 2 gR
(c)
gMv 2 2R
(d)
( g −1) Mv 2 2R
4. In the figure shown, there are 10 cells each of emf e and internal resistance r. The current through resistance R is (a) zero (b) e/r (c) 3e/r (d) 4e/r 8
Physics for you | november ‘15
5. A rectangular loop with a sliding connector of length 1 m is situated in a uniform magnetic field of 2 T perpendicular to the plane of loop. Resistance of connector is 2 W. Two resistances of 6 W and 3 W are connected as shown in the figure. The external force required to keep the connector moving with a constant velocity 2 m s–1 is (a) 2 N (b) 1 N (c) 4 N (d) 6 N 6. A thin lens of refractive index 1.5 and focal length in air 20 cm is placed inside a large container containing two immiscible liquids as shown in figure. If an object is placed at an infinite distance close to principal axis, the distance between two images will be (a) 25 cm (b) 40 cm (c) 65 cm (d) 85 cm Solution Senders of Physics Musing 1. 2. 3. 4.
set-27 Manmohan Krishna (Bihar) Anubhav Jana (WB) Shiekh Md. Shakeel Hassan (Assam) Swati Shah (Rajasthan)
1. 2. 3. 4.
set-26 Md. Samim Jahin (Assam) Deep Anand Basumatary (Assam) Harsimran Singh (Punjab) Sayantan Bhanja (WB)
7. The figure shows several equipotential lines. Comparing between points A and B, choose the best possible statement. (a) The electric field has a greater magnitude at point A and is directed to left. (b) The electric field has a greater magnitude at point A and is directed to right. (c) The electric field has a greater magnitude at point B and is directed to left. (d) The electric field has a greater magnitude at point B and is directed to right. 8. A light wire AB of length 10 cm can slide on a vertical frame as shown in figure. There is a film of soap solution trapped between the frame and the wire. Find the mass of the load W that should be suspended from the wire to keep it in equilibrium. Neglect friction. Surface tension of soap solution is 25 dyne cm–1. (Take g = 10 m s–2) (a) 0.25 g (b) 0.50 g (c) 2.50 g (d) 5.00 g
ParagraPh tyPe
Read the given paragraph and answer question number 9 and 10. Consider the situation shown in figure in which a block A of mass 2 kg is placed over a block B of mass 4 kg. The combination of the blocks are placed on an inclined plane of inclination 37° with horizontal. The system is released from rest. (Take g = 10 m s–2 and sin 37° = 0.6) 9. The coefficient of friction between block B and inclined plane is 0.4 and in between the two blocks is 0.5. Then (a) Both blocks will move but block A will slide over the blocks B. (b) Both blocks will move together. (c) None of them will move. (d) Only block A will move. 10. The frictional force acting between the blocks will be (a) 8 N (b) 6.4 N (c) 4 N (d) zero
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Physics for you | november ‘15
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one or more oPtions correct tyPe questions
1. Assume ABCDEF to be a regular hexagon. Choose the correct E D statements. + DB + BE = 0 (a) ED F C FE = (b) BC = 2FE (c) AD A B (d) DC = − AF 2. A spring mass system is hanging from the ceiling of an elevator in equilibrium as shown in figure. The elevator suddenly starts accelerating upwards with acceleration a, consider all the statements in the reference frame of elevator and choose the correct one(s). (a) The frequency of oscillation is
k
m
1 k . 2π m
ma . k m ( g + a) (c) The amplitude of resulting SHM is . k (d) Maximum speed of block during oscillation is m k a . (b) The amplitude of the resulting SHM is
3. An ideal gas has molar heat capacity at constant pressure CP = 5R . The gas is 2 kept in a cylindrical vessel fitted with a piston which is free to move. Mass of the frictionless piston is 9 kg. Initial volume of the gas is 0.0027 m3 and cross-section area of the piston is 0.09 m2. The initial temperature of the gas is 300 K. 12
Physics for you | NOVember ‘15
class-Xi
Atmospheric pressure P0 = 1.05 × 105 N m–2. An amount of 2.5 × 104 J of heat energy is supplied to the gas, then (a) Initial pressure of the gas is 1.06 × 105 N m–2. (b) Final temperature of the gas is 1000 K. (c) Final pressure of the gas is 1.06 × 105 N m–2. (d) Work done by gas is 9.94 × 103 J. 4. A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in figure. Assume both the friends d apply forces of equal magnitude. Choose the correct statements. (a) The force exerted by both the friends decreases as the man moves up. (b) T h e f o r c e ap p l i e d by e a c h f r i e n d i s mg 2 d + 4h2 when the man is at depth h. 4h (c) The force exerted by both the friends increases as the man moves up. mg 2 d + h2 (d) The force applied by each friend is h when the man is at depth h. 5. Two balls are thrown from an inclined plane at angle of projection a with the plane, one up the incline and other down the incline as shown in figure. (Here, T stands for total time of flight).
Which of the following are correct? v 2 sin2 α (a) h1 = h2 = 0 2 g cos q 2v0 sin α (b) T1 = T2 = g cos q (c) R2 – R1 = g(sinq) T12 (d) vt = vt 1
2
6. Two identical buggies move one after other due to inertia (without friction) with the same velocity v0. A man of mass m rides the rear buggy. At a certain moment, the man jumps into the front buggy with a velocity u relative to his buggy. If mass of each buggy is equal to M and velocity of buggies after jumping of man are vrear and vfront. Then m (a) vrear = v0 + u m+M m u (b) vrear = v0 − m+M mM (c) vfront = v0 + u (m + M )2 mM u (d) vfront = v0 − (m + M )2 7. A spherical body of radius R rolls on a horizontal surface with linear velocity v. Let L1 and L2 be the magnitudes of angular momenta of the body about centre of mass and point of contact P respectively. Then (here K is the radius of gyration about its geometrical axis) (a) L2 = 2L1 if radius of gyration K = R (b) L2 = 2L1 for all cases (c) L2 > 2L1 if radius of gyration K < R (d) L2 > 2L1 if radius of gyration K > R 8. Two solid spheres A and B of equal volumes but of different densities dA and dB are connected by a string. They are fully immersed in a fluid of density dF. They get arranged A into an equilibrium state as shown in the figure with a tension in the string. B The arrangement is possible only if (a) dA < dF (b) dB > dF (c) dA > dF (d) dA + dB = 2dF 9. A body of mass m is attached to a spring of spring constant k which hangs from the ceiling of an elevator at rest in equilibrium. Now the elevator starts accelerating upwards with its acceleration varying with time as a = pt + q, where p and q are
positive constants. In the frame of elevator, (a) The block will perform SHM for all value of p and q. (b) The block will not perform SHM in general for all value of p and q except p = 0. (c) The block will perform SHM provided for all value of p and q except p = 0. (d) The velocity of the block will vary simple harmonically for all value of p and q. 10. A string of mass m is fixed at both its ends. The fundamental mode of string is excited and it has an angular frequency w and the maximum displacement amplitude A. Then (a) The maximum kinetic energy of the string is 1 EK = mA2 w2 . 4 (b) The maximum kinetic energy of the string is 1 EK = mA2 w2 . 2 (c) The mean kinetic energy of the string averaged 1 over one periodic time is < EK > = mA2 w2 . 4 (d) The mean kinetic energy of the string averaged 1 over one periodic time is < EK > = mA2 w2 . 8 11. A bottle is kept on the ground as shown in the figure. The bottle can be modelled as having two cylindrical zones. The lower zone of the bottle has a cross-sectional radius of R 2 and is filled with honey of density 2r. The upper zone of the bottle is filled with the water of density r and has a crosssectional radius R. The height of the lower zone is H while that of the upper zone is 2H. If now the honey and the water parts are mixed together to form a homogeneous solution, then (Assume that total volume does not change) (a) The pressure inside the bottle at the base will remain unaltered. (b) The normal reaction on the bottle from the ground will remain unaltered. (c) The pressure inside the bottle at the base will 1 increase by an amount rgH . 2 (d) The pressure inside the bottle at the base will 1 decrease by an amount rgH . 4 Physics for you | November ‘15
13
12. A particle moving with kinetic energy 3 J makes an elastic collision (head-on) with a stationary particle which has twice its mass. During impact (a) The minimum kinetic energy of system is 1 J (b) The maximum elastic potential energy of the system is 2 J. (c) Momentum and total energy are conserved at every instant. (d) The ratio of kinetic energy to potential energy of the system first decreases and then increases. 13. Two blocks A and B each of mass m are connected by a massless spring of natural length L and spring constant k. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length as shown in figure. A third identical block C, moving on the floor with a speed v along the line joining A and B, collides with A. Then
(a) The maximum compression of the spring is v m/k . (b) The maximum compression of the spring is v m/2k . (c) The kinetic energy of A-B system at maximum compression of the spring is zero. (d) The kinetic energy of A-B system at maximum compression of the spring is mv2/4. 14. Three planets of same density and with radii R1, R2 and R3 such that R1 = 2R2 = 3R3, have gravitational fields on the surfaces E1, E2, E3 and escape velocities v1, v2, v3 respectively. Then E1 E1 1 = (a) (b) E = 3 E2 2 3 v1 1 v1 = =2 (c) (d) v3 3 v2 15. Water is flowing smoothly through a closed pipe system. At one point A, the speed of the water is 3.0 m s–1 while at another point B, 1.0 m higher, the speed is 4.0 m s–1. The pressure at A is 20 kPa when the water is flowing and 18 kPa when the water flow stops. Then (a) the pressure at B when water is flowing is 6.7 kPa. (b) the pressure at B when water is flowing is 8.2 kPa. (c) the pressure at B when water stops flowing is 10.2 kPa. (d) the pressure at B when water stops flowing is 8.2 kPa. 14
Physics for you | NOVember ‘15
solutions
1. (a, b,c, a regular d):As ABCDEF is hexagon, \ FE = BC , AD = 2FE , DC = − AF Also, from triangle law addition, of vector ED + DB + BE = 0. 2. (a, b, d) : As it is a isochronous system 1 k \ υ= 2π m ma From the reference frame of elevator, A = k k ma m vmax = wA = = a m k k 5R 3R , CV = 2 2 3 2 ∆W n(CP − CV ) = = 1− = 5 5 ∆Q nC p
3. (a, c) : CP =
2∆Q 2 .5 = 2× × 104 = 104 J 5 5 Pressure is constant and equal to mg 9 × 10 P = P0 + = 1.05 × 105 + A 0.09 = 1.06 × 105 N m–2 T 4. (b, c) : From figure, h sin q = d2 d/2 h2 + 4 mg As man moves slowly 2T sin q = mg mg T= 2sin q As man moves upward, q becomes small \ sin q decreases ⇒ T increases \ ∆W =
mg T= 2×h
T h
2 2 d 2 mg d + 4h h + = 4 4h 2
5. (a, b, c) : Maximum height of projectile on an (v sin a)2 inclined plane, h1 max = 0 = h2 max 2 g cos q ⇒ (a) is correct Time of flight 2v sin a T1 = 0 = T2 ⇒ (b) is correct g cos q
where a = angle of projection from inclined plane q = angle of inclination of surface. 1 R1 = (v0 cos a)T1 − g sin q T12 2 (Range upward the inclined plane) 1 R2 = (v0 cos a)T2 + g sin q T22 2 (Range downward the inclined plane) ⇒ (R2 – R1) = g sin q T12 ⇒ (c) is correct vt and vt are the velocities of the particles at their 1 2 maximum height. Let the particles reach their maximum heights at time t1 and t2 respectively. Hence, 0 = (v0 sin a) – (g cos q) t1 v sin a ⇒ t1 = 0 g cos q v sin a Similarly, t2 = 0 . g cos q Hence, t2 = t1 Hence, vt = v0 cos a – (g sin q) t1 1 vt = v0 cos a + (g sin q) t2 2
⇒ vt ≠ vt 1 2
6. (b, c) : As no external force is applied to the system v0
v
v0
u+v
v0
Conserving linear momentum of man and rear buggy, (M + m)v0 = Mv + m(v + u) mu = vrear ⇒ v = v0 − M +m Conserving linear momentum of man and front buggy, m(u + v) + Mv0 = (M + m)v′ mu m u + v0 − + Mv0 = (M + m)v ′ M + m Mmu = ( M + m)v ′ M +m Mmu v ′ = v0 + = vfront ( M + m)2 7. (a, d) 8. (a, b, d) : Let V be the volume of each sphere and T is the tension in the string. For the string to be taut, dFVg > dAVg or dF > dA (a) is correct and dBVg > dFVg or dB > dF ( M + m)v0 +
dFVg
(b) is correct For an equilibrium dFVg + dFVg + T = T + dAVg + dBVg or dA + dB = 2dF (d) is correct 9. (c, d) : In the frame of elevator d2x mg + ma − kx = m dt 2 2 d x k m( g + a) ⇒ = − x − 2 m k dt or
d2x dt
2
=−
k m( g + pt + q) x− m k
There is a term involving t on R.H.S., this does not represent S.H.M. unless p = 0 Differentiating with respect to time d 2v k mp d3x k dx mp = − − or = − v − 3 2 m dt k m k dt dt Thus the velocity of the block will vary simple harmonically. 10. (a, d) : Let the displacement of the string be given by πx y( x , t ) = A sin cos(wt + d) L where d is a phase factor. So the transverse velocity is given by πx ∂y v( x , t ) = = − wA sin sin(wt + d) ∂y L The maximum kinetic energy is equal to the string’s total energy of oscillation. Note that all points of the string achieve their maximum kinetic energy at the same instant of time, where y = 0 for all x. Since m dm = mdx where m = is the mass per unit L length of the uniform string. The maximum kinetic energy, 1 ∂y 2 EK = max imum of ∫ dm 2 ∂t 1 L ∂y 2 = maximum of ∫ m dx 2 0 ∂y 2
A T
dAVg
T
dFVg B dBVg
∂y The maximum value of , occurs when ∂y 2 sin (wt + d) = 1 L m πx Hence EK = A2 w2 ∫ sin2 dx L 2 0
Physics for you | NOVember ‘15
15
L
πx The integral ∫ sin dx over the half-cycle has L 0 L the average value of 2 1 mL 1 Hence, EK = A2 w2 = mA2 w2 2 4 2 2
\ (a) is correct The mean kinetic energy of the string averaged over one periodic time is obtained by integrating the time dependent factor sin2 (wt + d) over one period, 0 to T. Now since
T
∫ sin
2
(wt + d) dt =
0
T 2
The mean kinetic energy of the string averaged over one periodic time is T
E E < EK > = K ∫ sin2 (wt + d) dt = K T 2 0
1 1 1 mA2 w2 = mA2 w2 2 4 8 (d) is correct =
\
11. (b, c) : Initial pressure at the bottom = rg × 2H + 2r × g × H = 4rgH Final density of the homogeneous mixture r × A × 2H + 2r × 2 A × H 3 = = r A × 2H + 2 A × H 2 Final pressure at the bottom =
3 9 r × g × 3H = rgH 2 2
12. (a, b, c, d) : In a head on elastic collision between two particles, the kinetic energy becomes minimum and potential energy becomes maximum at the instant when they move with a common velocity. The momentum and energy are conserved at every instant. Let m and u be the mass and initial velocity of the first particle, 2m be the mass of second particle and v be the common velocity. u 1 Then, mu2 = 3 J ; mu = (m + 2m) v or v = 3 2 Minimum kinetic energy of system 2
1 u = (3m) = 1 J 2 3 Maximum potential energy of system = 2 J 13. (b, d) : After collision of C with A, let velocity acquired by A and B be v′ and spring gets compressed by length x. Using law of conservation of linear momentum, we have 16
Physics for you | NOVember ‘15
mv = mv′ + mv′ or v′ = v/2 Using law of conservation of mechanical energy, we have 1 2 1 1 1 mv = mv ′2 + mv ′2 + kx 2 2 2 2 2 2 2 v v 2 2 or mv = m + m + kx 2 2 mv 2 = kx 2 or 2 1/ 2 m or x = v \ (b) is correct 2k At maximum compression of the spring, the kinetic energy of A-B system will be 2
1 1 mv 2 v = mv ′2 + mv ′2 = mv 2 = m = 2 2 2 4 \ (d) is correct 4 G πR3r GM 3 14. (b, c) : E = 2 = 2 R R or E∝R E1 R1 2R2 = = = 2 \ (a) is not correct. E2 R2 R2 E1 R1 3R3 = = = 3 \ (b) is correct. E3 R3 R3 Escape velocity, v =
2GM = R
=
2G 4 3 πR r R 3
8 2 πR r G or v ∝ R 3
v1 R1 = = 2 \ (c) is correct. v2 R2
v1 R1 = = 3 \ (d) is not correct. v3 R3 15. (a, d) : Let P1, h1, and v1 and P2, h2 and v2 represent the pressures, heights and velocities of flow at the two points respectively. According to the Bernoulli’s theorem 1 1 P1 + rgh1 + rv12 = P2 + rgh2 + rv22 ...(i) 2 2 Putting v1 = 3.0 m s–1, v2 = 4.0 m s–1, (h2 – h1) = 1 m, P1 = 20 kPa we get, 10 3 P2 = 20 + 10 3 × 9.8 ( −1) + [9 − 16] × 10 −3 2
= 20 – 9.8 – 3.5 = 6.7 kPa Also when the flow stops, v1 = v2 = 0 and then from (i), P2 = 18 – 9.8 = 8.2 kPa
Physics for you | NOVember ‘15
17
Pressure inside a liquid
We choose atmospheric pressure = P0 At a depth h below the free surface, pressure = P. P0
h P
To find P, we choose a liquid column of height h and crosssectional area A. Since the liquid column is unaccelerated, P0A + mg = PA ⇒ P = P0 +
mg r( Ah) = P0 + g A A
⇒ P = P0 + rgh The additional pressure with respect to atmospheric pressure is known as Gauge pressure. Hence we conclude that pressure changes by an amount rgh on moving through a distance h vertically. Note that this result has been derived from equilibrium of the liquid column. Hence if the container or liquid was vertically accelerated, it would not be applicable. In such cases if the container is vertically accelerated, say upward with a, then (Fnet)upward direction = ma ⇒ (PA) – (P0A + mg) = ma m ⇒ P = P0 + ( g + a) A ⇒ P = P0 + r(g + a)h \ We replace g with geff where, g eff = g + (−a ) \ P = P0 + rgeff h So, it is interesting to see that we can also have a situation that all points inside a liquid irrespective of
their location, will have same pressure as atmospheric if geff = 0, as in case of free fall. Measurement of atmospheric pressure (P0)
We take a tub filled partially with mercury and a tube completely filled with mercury. We seal the mouth of the tube and invert it upside down with the mouth inside the mercury in the tub. The liquid in the tube drops down a little, creating almost vacuum in the upper closed end of the tube as shown. At equilibrium, PA + rgh = PB = PC = P0 ⇒ rgh = P0 [ PA = 0] Hence, measuring the length of the liquid column in the tube, P0 can easily be calculated. Taking P0 = 1 atm ≈ 105 N m–2 For Hg, r = 13.6 g cm–3, h comes out to be almost 760 mm or 76 cm. Hence you would often see that pressure is given in terms of length and not N m–2 or pascal. If instead of Hg, some other liquid is used, to find the height of liquid risen we can easily use, r1h1 = r2h2 Supposedly, we keep this set-up in an upward accelerating frame, then how will h change? Clearly, in such case also, atmospheric pressure does not change, we need to change g with geff. \ P1 = rgh = rg eff h′ = r(g + a)h′ g \ h′ = h g + a \ h′ < h ⇒ height of liquid column decreases.
Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata
18
Physics for you | november ‘15
Pressure difference in a horizontally accelerated container
In such case, clearly, the pressure at same horizontal level would not be same. To know the exact relation we consider a thin horizontal liquid column of length l as drawn. Hence from free body diagram,
a = gsin
n
gsi
g
gcos
lel
ral pa
e lin
nc oi
t
gcos = geff
Since, geff is perpendicular to inclined plane, hence free surface of liquid is parallel to inclined plane. Hence a = 0° U-tubes
(P2A – P1A) = ma ⇒ (P2 – P1)A = (rAl)a ⇒ P2 = P1 + ral \ DP = ral along horizontal, as DP = rgh along vertical. Vertically pressure increases in the direction of gravity, horizontally acceleration increases opposite to the direction of acceleration. Let us calculate the inclination of free surface with horizontal now. We can find pressure at B from A as well as C. PB = PA + rgh = PC + raL h a ⇒ tan q = = L g This could also have been found out using the fact that liquids cannot tolerate tangential force on its surface. Hence the free surface should be perpendicular to g eff as below. g eff = g + (−a ) [We revert the acceleration of container and add it vectorially to g ]
Let us apply this to a more complicated situation. Assume a container falling down on a smooth inclined plane. We have to find a. Clearly, a = gsinq downwards Hence, g eff = g + (−a )
Let us consider a U-tube filled with two immiscible liquids on two limbs. Note that, PC = PF but PA ≠ PD PB ≠ PE, even though (A, D) and (B, E) pair of points are at same level. Since if we move up from C to B and F to E, the change in height is same but we have different density of liquids hence r1h ≠ r2h. If difference in height of free surface of liquids is to be calculated in terms of H, PC = PF r2 ⇒ P0 + r1gh1 = P0 + r2 gH ⇒ h1 = H r1 r \ Dh = 1 − 2 H r1 Archimedes’ principle
If an object is submerged inside a liquid, partially or completely, it experiences an upward force by the liquid due to pressure difference along the vertical column of the liquid, which is equal to the weight of liquid displaced by the object. To prove this, let us imagine an object of cross-sectional area A and height H partially submerged till height h as shown here.
h l
s
H
FBD of object P0A
mg (P0 + l gh )A
\ By definition, force of upthrust (also known as buoyant force) is Fup = (P0 + rl gh)A – P0A = rl (hA)g = rlVsub g = weight of displaced liquid Physics for you | november ‘15
19
Note : This result is applicable only if the container is vertically unaccelerated, else we need to replace g with geff in the result. Now, let us see what would the condition of floatation for the object be. For equilibrium, Fup = mg ⇒ rl(Ah)g = rs(AH)g h rs ⇒ = ≤ 1 [ h ≤ H for floatation] H rl \ If rs ≤ rl, the object floats, else it sinks. Hence it does not matter how heavy an object is for floating, what matters is how dense the object is! h is Note here, that the fraction of submerged portion, H independent of g, hence even in accelerated containers, this same fraction will be submerged. Now, let us seen an application. Suppose a helium filled balloon is floating in air (their densities He air given as rHe and rair) with a string tied to a box as shown here. Now, if the box is accelerated towards right with acceleration a, we have to find the direction and angle with the vertical in which the string gets deflected at equilibrium. One would be tempted to say that the string will deflect towards left due to the pseudo force. But there is a basic point, one is missing here. As we saw force of upthrust being generated due to pressure difference along vertical column of liquid, similar to it a side thrust can also be generated if pressure difference is created along horizontal column and on similar approach it can be proved. Fside thrust = rl Vsub a Hence, since, rair > rHe, so side thrust will be greater than the pseudo force, so the balloon deflects towards right. airVg
HeVa
T
20
(air– He)Vg
airVa
HeVg
(air– He)Va T
Physics for you | november ‘15
\ tan q =
(rair − rHe )Va a = (rair − rHe )Vg g
Force exerted by a liquid on a vertical surface
Let us assume, that the wall of a dam of width w stops a liquid of height h from flowing. It obviously is experiencing force due to the liquid’s pressure. To find this, let us consider a horizontal strip of height dy at a depth y below the free surface. The force experienced by this surface due to pressure of liquid is dF = (rgy)(dyw) (P0 is not considered since it is due to atmosphere) h
⇒ F = ∫ dF = ∫ wrgydy 0
rgh2w = 2 \ Net horizontal force on vertical surface h = rg (hw ) 2 = pressure at centroid of submerged portion × area of submerged portion Force exerted by liquid on a horizontal surface
Let us consider a horizontal face of area A at a depth h below the free surface. We have to find the force exerted by the liquid only. Net vertical force = PA = (rgh)A = r(Ah)g = weight of liquid column above its surface nn
Solution Set-27
1. (d) : Here, d = 0.5 mm = 0.5 × 10–3 m D = 0.5 m l = 500 nm = 500 × 10–9 m The distance of third maxima from the second minima on the other side is 9 = b (where b is the fringe width) 2 9 lD 9 × 500 × 10−9 × 0.5 = = 2 d 2 × 0.5 × 10−3 = 2.25 × 10–3 m = 2.25 mm 2. (c) : During motion of the particle, total mechanical energy remains constant. At the surface of earth, total mechanical energy is GmM 1 2 Ei = − + mv0 R 2 GM 1 2 = − 2 mR + mv0 2 R 1 2 GM = − gmR + mv0 g = 2 2 R Total mechanical energy at height h = R is GmM 1 2 gmR 1 2 Ef = − + mv = − + mv 2R 2 2 2 According to law of conservation of mechanical energy, Ei = Ef 1 2 gmR 1 2 ∴ − gmR + mv0 = − + mv 2 2 2 or − 2 gR + v02 = − gR + v 2 ∴ v = v02 − gR 3. (a) : If we consider the cylindrical surface to be a ring of radius R, there will be an induced emf due to changing field. df dB E ∫ ⋅ dl = − dt = − A dt dB dB R dB ⇒ E(2 πR) = − A = − πR2 ⇒ E=− dt dt 2 dt ∴ Force on the electron eR dB F = − Ee = 2 dt 1 eR dB ⇒ Acceleration = 2 m dt As the field is increasing being directed inside the paper, hence there will be anticlockwise induced current (in order to oppose the cause) in the ring
(assumed). Hence there will be a force towards left on the electron. 4. (c) : Total time of flight is T = 4 s and if u is its initial speed and q is the angle of projection. Then 2u sinq T= =4 g or usinq = 2g ...(i) After 1 s velocity vector makes an angle of 45° with horizontal i.e., vx = vy ucosq = usinq – gt ( t = 1 s) ucosq = usinq – g ucosq = 2g – g (Using (i)) or ucosq = g ...(ii) Squaring and adding (i) and (ii), we get u2sin2q + u2cos2q = (2g)2 + (g)2 u2 = 5g2 = 5(10)2 m2 s–2 ∴ u = 22.36 m s–1 Dividing (i) by (ii), we get, u sin q 2 g = =2 u cos q g tanq = 2 or q = tan–1 (2) l 5. (d) : Here, T = 2π ...(i) g When lift is accelerated upwards with acceleration T a, let time period becomes . Then 2 T l ...(ii) = 2π 2 g +a Dividing Eq. (i) by Eq. (ii), we get 1/2
g +a a = 1 + g g Squaring both sides, we get a 4 = 1 + or a = 3g g 6. (c) : Lengths of the two inclined planes are h h l1 = and l2 = sin q1 sin q2 Accelerations of the block down the two planes are a1 = g sinq1 and a2 = g sinq2 1 1 As l1 = a1t12 and l2 = a2t22 2 2 l1 a1t12 t 22 a1l2 g sin q1 sin q1 ∴ = or 2 = = × l2 a2t22 t1 a2l1 g sin q2 sin q2 t sin q1 ∴ 2= t1 sin q2 7. (a) : Given : f = at2 + bt The magnitude of induced emf is 2=
Physics for you | November ‘15
21
a sin q 2 a a + cos q 2 a sin q tan = 2 (2 + cos q) a 2 tan 2 tan a = 2a 1 − tan 2 sin q 2⋅ sin q 2 + cos q = 1 + cos q sin2 q 1− (2 + cos q)2 1 On solving, cos q = − 2 ∴ q = 120°
df d = (at 2 + bt ) = 2at + b dt dt Current flowing, I = | ε | = 2at + b R R ε=
τ
Average emf =
∫ εdt 0 τ
a tan = 2
τ
=
∫ (2at + b)dt 0
∫ dt
τ
= aτ + b
0
Total charge flowing, τ
τ
aτ2 + bτ (2at + b) dt = R R 0
q = ∫ Idt = ∫ 0
T 8. (a) : Free body diagram for m For m, a m N mg – T = m 2a … (i) 2a N = ma … (ii) mg Free body diagram for M For M 2T – N = Ma …(iii) 2mg On solving, a = ( M + 5m) ∴ Net acceleration of m, 2 5mg am = 4a2 + a2 = 5a = (5m + M ) B sin q 9. (b) : tan a = ...(i) A + B cos q where a is the angle made by the vector ( A + B) with A. B sin q Similarly, tan b = ...(ii) A − B cos q where b is the angle made by the vector ( A − B) with A. Note that the angle between A and (− B) is (180° – q). Adding (i) and (ii), we get B sin q B sin q tan a + tan b = + A + B cos q A − B cos q 2 AB sin q − B sin q cos q + AB sin q + B2 sin q cos q = ( A + B cos q)( A − B cos q) 2 AB sin q = 2 ( A − B2 cos2 q)
10. (a) :
1 4 5
8
10
13
26
S
22
Physics for you | November ‘15
…(i)
D R Y
I T
I C E
R Y
A
Y
O
N
E M
T I 27G N A L 28 M A C H I N E R T 29
Y D M I R R
7
N D F A R M S C M 9
12
15
D A T
A
F Y D R T 16 F E S 20 O N L 22 R S C L E R C I O 14
E
E
N W I N U C L E A R I L A 11
3
K
6
E
H
E
2
S
C K A E
Z
T
F L
R A I V 19 18 P O S P R U L N
B sin q A + B cos q sin q tan a = . 1 + cos q
nn
solution of october 2015 crossword
tan a =
…(ii)
L
Y O C T O T I
P
F
L
O P
Y C E L L P 17
W E I G H T O F 23 O M E 24T E R O 25A S I
21
S
L
U N
K
R E
L Y F W
M C
I
U A E Y
C L O U D L S R
A H L O I
I
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Winners (October 2015) Basheer Mazahar (Varanasi) sidharth sankar sahu (Odisha) Ayushi Tripathi (Delhi) solution senders (september 2015) Anoop Jain (Delhi) sanchit Mehta (Haryana) Viraj Thapa (Assam)
chapterwise McQs for practice
Useful for All National and State Level Medical/Engg. Entrance Exams
Mechanical ProPerties of fluids
1. A hemispherical bowl just floats without sinking in a liquid of density 1.2 × 103 kg m–3. If outer diameter and the density of the bowl are 1 m and 2 × 104 kg m–3 respectively, then the inner diameter of the bowl will be (a) 0.94 m (b) 0.96 m (c) 0.98 m (d) 0.99 m 2. A body of density r is dropped from rest at a height h into a lake of density s, where s > r. Neglecting all dissipative forces, calculate the maximum depth to which the body sinks before returning to float on the surface. h hr (a) s − r (b) s hr hs (c) s − r (d) s−r 3. Water in a vessel of uniform cross-section escapes through a narrow tube at the base of the vessel. Which of the following graphs represents the variation of the height h of the liquid with time t? h
h
(a)
13.6 g cm–3 and the angle of contact of mercury and water are 135° and 0° respectively, the ratio of surface tension of water and mercury is (a) 1 : 0.15 (b) 1 : 3 (c) 1 : 6.5 (d) 1.5 : 1 5. A metallic sphere of mass M falls through glycerine with a terminal velocity v. If we drop a ball of mass 8M of same metal into a column of glycerine, the terminal velocity of the ball will be (a) 2 v (b) 4 v (c) 8 v (d) 16 v 6. A cylindrical drum, open at the top, contains 15 L of water. It drains out through a small opening at the bottom. 5 L of water comes out in time t1, the next 5 L in further time t2 and the last 5 L in further time t3. Then (a) t1 < t2 < t3 (b) t1 > t2 > t3 (c) t1 = t2 = t3 (d) t2 > t1 = t3 7. A sealed tank containing a liquid of density r moves with a horizontal acceleration a, as shown in figure. The difference in pressure between the points A and B is l
(b) h
t
(c)
h
t
(d) t
t
4. Water rises to a height of 10 cm in a capillary tube and mercury falls to a depth of 3.42 cm in the same capillary tube. If the density of mercury is
h
A
a
B
(a) h r g + l r a (c) h r g
(b) h r g – l r a (d) l r a
8. The surface area of air bubble increases four times when it rises from bottom to top of a water tank where the temperature is uniform. If the Physics for you | november ‘15
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atmospheric pressure is 10 m of water, the depth of the water in the tank is (a) 30 m (b) 40 m (c) 70 m (d) 80 m
(c) On the surface of the moon, the height is more than h. (d) In a lift moving down with constant acceleration, height is less than h.
4 th of its 5 volume submerged, but it just floats in a liquid. What is the density of liquid? (a) 750 kg m–3 (b) 800 kg m–3 (c) 1000 kg m–3 (d) 1250 kg m–3
14. A candle of diameter d is floating on a liquid in a cylindrical container of diameter D(D>>d) as shown in figure. If it is burning at the rate of 2 cm h–1, then the top of the candle will
9. A block of wood floats in water with
10. A spherical ball is dropped in a long column of viscous liquid. Which of the following graphs represent the variation of (i) gravitational force with time (ii) viscous force with time (iii) net force acting on the ball with time? F Q R
(a) Q, R, P (c) P, Q, R
P
t
(b) R, Q, P (d) R, P, Q
11. Water is flowing through a horizontal pipe. If at one point pressure is 2 cm of Hg and velocity of flow of the liquid is 32 cm s–1 and at another point, velocity of flow is 40 cm s–1, the pressure at this point is (a) 1.45 cm of Hg (b) 1.98 cm of Hg (c) 1.67 cm of Hg (d) 1.34 cm of Hg 12. The rate of flow of glycerine of density 1.25 × 103 kg m–3 through the conical section of a pipe, if the radii of its ends are 0.1 m and 0.04 m and the pressure drop across its length is 10 N m–2, is (a) 5.28 × 10–4 m3 s–1 (b) 6.28 × 10–4 m3 s–1 (c) 7.28 × 10–4 m3 s–1 (d) 8.28 × 10–4 m3 s–1 13. Water rises in a capillary tube to a height h. Choose the false statement regarding a capillary rise from the following. (a) On the surface of Jupiter, height will be less than h. (b) In a lift, moving up with constant acceleration, height is less than h. 24
Physics for you | november ‘15
L L d D
(a) (b) (c) (d)
remain at the same height fall at the rate of 1 cm h–1 fall at the rate of 2 cm h–1 go up at the rate of 1 cm h–1
15. A frame made of metallic wire enclosing a surface area A is covered with a soap film. If the area of the frame of metallic wire is reduced by 50%, the energy of the soap film will be changed by (a) 100% (b) 75% (c) 50% (d) 25% therMal ProPerties of Matter
16. The plots of intensity versus wavelength for three black bodies at temperatures T1, T2 and T3 respectively are as shown. Their temperatures are such that I
T3 T1
T2
(a) T1 > T2 > T3 (c) T2 > T3 > T1
(b) T1 > T3 > T2 (d) T3 > T2 > T1
17. A solid copper sphere (density r and specific heat capacity c) of radius r at an initial temperature 200 K is suspended inside a chamber whose walls are at almost 0 K. The time required (in ms) for the temperature of the sphere to reach 100 K is 7 r rc 80 r rc (a) (b) 80 s 7 s 7 r rc 7 r rc (c) (d) 27 s 27 s
18. A clock with an iron pendulum keeps correct time at 15°C. What will be the error in time per day, if the room temperature is 20°C? (The coefficient of linear expansion of iron is 0.000012°C–1.) (a) 2.6 s (b) 6.2 s (c) 1.3 s (d) 3.1 s 19. A body cools from 60°C to 50°C in 10 min. If room temperature is 25°C, temperature of body at the end of next 10 min will be (a) 38.5°C (b) 40°C (c) 45°C (d) 42.8°C 20. Two rods of same length and material transfer a given amount of heat in 12 s, when they are joined end to end (i.e., in series). But when they are joined in parallel, they will transfer same heat under same conditions in (a) 24 s (b) 3 s (c) 48 s (d) 1.5 s 21. A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt should be (a) 450 (b) 900 (c) 225 (d) 1800 22. A body in laboratory takes 4 min to cool from 61°C to 59°C. If the laboratory temperature is 30°C, then the time taken by it to cool from 51°C to 49°C is (a) 4 min (b) 6 min (c) 8 min (d) 5 min 23. The wavelength of maximum intensity of radiation emitted by a star is 289.8 nm. The radiation intensity for the star is (Take s = 5.67 × 10–8 W m–2 K–4, Wien’s constant, b = 2898 mm K) (a) 5.67 × 108 W m–2 (b) 5.67 × 1012 W m–2 (c) 5.67 × 107 W m–2 (d) 5.67 × 1014 W m–2 24. The reading of Centigrade thermometer coincides with that of Fahrenheit thermometer in a liquid. The temperature of the liquid is (a) –40°C (b) 0°C (c) 100°C (d) 300°C 25. For a black body at temperature 727 °C, its radiating power is 60 W and temperature of surrounding
is 227°C. If the temperature of the black body is changed to 1227°C, then its radiating power will be (a) 120 W (b) 240 W (c) 304 W (d) 320 W 26. A metal plate 4 mm thick has a temperature difference of 32°C between its faces. It transmits 200 kcal h–1 through an area of 5 cm2. Thermal conductivity of the material is (a) 58.33 W m–1 °C–1 (b) 33.58 W m–1 °C–1 (c) 5 × 10–4 W m–1 °C–1 (d) 5 × 10–2 W m–1 °C–1 27. A 2 kg copper block is heated to 500°C and then it is placed on a large block of ice at 0°C. If the specific heat capacity of copper is 400 J kg–1°C–1 and latent heat of fusion of water is 3.5 × 105 J kg–1, the amount of ice that can melt is (a) (7/8) kg (b) (7/5) kg (c) (8/7) kg (d) (5/7) kg 28. The maximum wavelength of radiation emitted at 2000 K is 4 mm. What will be the maximum wavelength emitted at 2400 K? (a) 3.3 mm (b) 0.66 mm (c) 1 m (d) 1 mm 29. The net rate at which heat is lost by a body due to radiation does not depend upon (a) temperature of the body (b) temperature of the surroundings (c) material of the body (d) nature of its surface 30. We plot a graph, having temperature in °C on x-axis and in °F on y-axis. If the graph is straight line, then it (a) passes through origin (b) intercepts the positive x-axis (c) intercepts the positive y-axis (d) intercepts the negative axis of both x-and y-axis solutions 1. (c) : L e t D 1 b e t h e i n n e r d i a m e t e r of t h e hemispherical bowl. As bowl is just floating so 3 3 3 4 1 D1 3 4 1 p × 1.2 × 10 = p − × (2 × 104 ) 3 2 3 2 2
Physics for you | november ‘15
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or ⇒
1.2 × 103 = 1 − D13 2 × 10 4 1/3 1/3 1. 2 18.8 D1 = 1 − = 20 20
On solving, we get D1 = 0.98 m 2. (c) : The speed of the body just before entering the liquid is u = 2 gh . The buoyant force FB of the lake, i.e., upward thrust of liquid on the body is greater than the weight of the body W, since s > r. If V is the volume of the body and a is the acceleration of the body inside the liquid, then FB – W = ma or sVg – rVg = rVa or (s – r)g = ra (s − r)g or a = r Using the relation, v2 = u2 + 2as, we have (s − r) 0 = ( 2 gh )2 − 2 g s r hr s−r 3. (a) : Let dV be the decrease in volume of water in vessel in time dt. Therefore rate of decrease of water in vessel = rate of water flowing out of narrow tube or
s=
So, −
4 dV p(P1 − P2 )r = 8 hl dt
But, P1 – P2 = hrg \
−
dV p(hrg )r 4 (prgr 4 ) = = × (h × A) 8 hl 8 hl × A dt
where h × A = volume of water in vessel at a time t =V prgr 4 × V dt = − lV dt \ dV = − 8hlA dV or = − l dt V prgr 4 = l = constant where, 8hlA Integrating it within the limits, as time changes from 0 to t, volume changes from V0 to V. V = − lt or ln V0 or V = V0–lt where, V0 = initial volume of water in vessel = Ah0 Therefore,
26
Physics for you | november ‘15
h = h0e–lt Thus, the variation of h and t will be represented by exponential curve as given by (a). 2S cos θ 4. (c) : As, h = rrg hr rg hr or S ∝ \ S= 2 cos θ cos θ Sw h1 cos θ2 r1 = × × \ SHg h2 cos θ1 r2 =
cos 135° 10 1 × × (−3.42) cos 0° 13.6
=
10 0.707 1 × = 3.42 13.6 6.5
4 3 4 pr r and 8M = pR 3r, 3 3 3 3 So, R = 8r ⇒ R = 2r
5. (b) : As, M =
2
2
v R 2r Now v ∝ r2 so, 1 = = = 4 r v r or v1 = 4 v 6. (a) : If h is the initial height of liquid in drum above the small opening, then velocity of efflux, v = 2 gh . As the water drains out, h decreases, hence v decreases. This reduces the rate of drainage of water. Due to which, as the draining continues, a longer time is required to drain out the same volume of water. So, clearly, t1 < t2 < t3. 7. (a) : Since points A and C are in the same horizontal line but separated by distance l and liquid tank is moving horizontally with acceleration a, hence PC – PA = l r a or PC = PA + l r a l h
C
A
a
B
Points B and C are vertically separated by h \ PB – PC = h r g or PB – (PA + l r a) = h r g or PB – PA = h r g + l r a 1/ 2
A 8. (c) : Surface area, A = 4pr2 or r = 4p 3/2
4 3 4 A pr = p = kA3/2 3 3 4p 4p 1 w here × = k = constant 3 (4 p)3/2
Volume V =
Using Boyle’s law, we have P1V1 = P2V2
(10 + h) kA13/ 2 PV or P2 = 1 1 = V2 kA23/ 2 A or P2 = (10 + h) 1 A2
3/2
1 = (10 + h) 4 =
As P2 = 10 m of water, so 10 + h 10 = or 80 = 10 + h 8 or h = 70 m 9. (b) : Density of water = Density of solid ×
3/2
10 + h 8
total volume of solid
immersed volume V 5 \ r1 = r × 4 = r 4 V 5 V And density of liquid, r2 = r = r V 5 \ r1 = r2 4 4 4 or r2 = r1 = × 1000 = 800 kg m −3 5 5 10. (c) : Gravitational force remains constant on the falling spherical ball. It is represented by straight line P. The viscous force (F = 6 phrv) increases as the velocity increases with time. Hence, it is represented by curve Q. Net force = gravitational force – viscous force. As viscous force increases, net force decreases and finally becomes zero. Then the body falls with a constant terminal velocity. It is thus represented by curve R. 11. (b) : As per Bernoulli’s theorem, 1 1 P1 + rv12 = P2 + rv22 2 2 1 2 or P1 − P2 = r (v2 − v12 ) 2 1 or P1 − P2 = × 1[402 − 322 ] = 288 dyne cm −2 2 288 = = 0.02 cm of Hg 13.6 × 980 \ P2 = P1 –0.02 = 2 – 0.02 = 1.98 cm of Hg 12. (b) : According to continuity equation, v2 A1 p × (0.1)2 25 ...(i) = = = 2 v1 A2 p × (0.04) 4
According to Bernoulli’s equation for horizontal tube, 1 1 P1 + rv12 = P2 + rv22 2 2 2(P1 − P2 ) 2 2 i.e., v2 − v1 = r (2 × 10) i.e., v22 − v12 = = 16 × 10−3 m2 s −2 ...(ii) 3 (1.25 × 10 ) Substituting the value of v2 from equation (i) in (ii)
(6.25v1)2 – v12 = 16 × 10–3, i.e., v1 = 0.02 m s–1 So rate of flow through the tube = A1v1 (= A2v2) = p × (0.1)2 × 0.02 = 6.28 × 10–4 m3 s–1 13. (d) 14. (b) : Volume of candle = area × length 2
d = p × 2L 2 Weight of candle = weight of liquid displaced Vrg = V′rg d2 d2 × 2L r = p × L r′ or p 4 4 r 1 ...(i) ⇒ = r′ 2 Since, candle is burning at the rate of 2 cm h–1, then after an hour, candle length is 2L – 2 \ (2L – 2)r = (L – x)r′ r L−x \ = r′ 2(L − 1) 1 L−x (using (i)) or = 2 2(L − 1) \ x = 1 cm Hence, it also decreases 1 cm outside, so, it falls at the rate of 1 cm h–1. 15. (c) : Surface energy = surface tension × surface area or U = S × 2A A New surface energy, U1 = S × 2 = S × A 2 % decrease in surface energy U − U1 2SA − SA = × 100% = × 100% = 50% U 2SA 1 16. (b) : According to Wien’s law, lm ∝ . T From the figure, (lm)1 < (lm)3 < (lm)2, therefore T1 > T3 > T2 Physics for you | november ‘15
27
dT sA 4 = (T − T04 ) dt mcJ Here, fall in temperature of body dT = (200 – 100) = 100 K, temperature of surrounding T0 = 0 K, initial temperature of body T = 200 K
17. (b) :
18.
19.
20.
21.
s 4 pr 2 100 (2004 − 04 ) = 4 3 dt pr rcJ 3 rrcJ rrc 4.2 × 10−6 s = ⋅ × 10−6 (As J = 4.2) \ dt = 48s s 48 7 r rc = ms 80 s (a) : Here, DT = 20 – 15 = 5°C a = 0.000012°C–1 = 12 × 10–6 °C–1 1 Time lost per day = a(DT ) × 86400 s 2 1 = × 12 × 10−6 × 5 × 86400 s 2 = 2.592 s ≈ 2.6 s (d) : According to Newton’s law of cooling T2 − T1 T + T = K 2 1 − T0 t 2 60 − 50 60 + 50 \ =K − 25 = 30 K 10 2 1 or K = 30 For next 10 min 50 − T T T 50 + T =K − 25 = K = 10 2 60 2 300 = 42.8° C or 70T = 3000, T = 7 DQ ( Dx ) (b) : Dt = KA (DT ) When two rods of same length are joined in parallel Dx A′ = 2A and Dx ′ = 2 Dt 1 = × 12 s = 3 s \ Dt ′ = 4 4 (d) : For a spherical black body of radius r at T K, Power radiated = energy radiated per second P = 4pr2 (sT4) \
2 4 P2 r2 T2 = P1 r12 T14 2
P2 1 = × (2)4 = 4 450 2 P2 = 4 × 450 = 1800 watt 28
Physics for you | november ‘15
22. (b) : According to Newton’s law of cooling T1 − T2 T + T = K 1 2 − T0 t 2 61 − 59 61 + 59 \ =K − 30 = 30 K 4 2 or 60 K = 1 ⇒ K = 1/60 51 − 49 51 + 49 =K − 30 = 20 K Now t 2 1 1 = 20 × = 60 3 \ t = 3 × 2 = 6 min 23. (a) : Here, lm = 289.8 nm = 289.8 × 10–9 m s = 5.67 × 10–8 W m–2 K–4 b = 2898 mm K = 2898 × 10–6 m K If T is the temperature of star, then according to Wien’s law, lm T = b
2898 × 10−6 b = = 104 − 9 lm 289.8 × 10 From Stefan’s law, E = sT4 = 5.67 × 10–8 × (104)4 = 5.67 × 108 W m–2 24. (a) : Temperature on Celsius scale and Fahrenheit scale are related as TC − 0 TF − 32 = 100 180 If the temperature is T at which the readings of two scales coincide, then from T T − 32 = \ T = −40° C 100 180 T=
25. (d) : Radiating power, P ∝ (T 4 − T0 4 ) Here, T is temperature of the black body and T0 is temperature of surrounding. P2 T24 − T04 P2 (1500)4 − (500)4 \ = \ = 4 4 P1 T1 − T0 60 (1000)4 − (500)4 80 or P2 = × 60 = 320 W 15 26. (a) : Here, Dx = 4 mm = 4 × 10–3 m, A = 5 cm2 = 5 × 10–4 m2, DT = 32°C Heat transmit per hour DQ = 200 kcal h −1 Dt 200 × 1000 × 4.2 −1 = J s = 233.33 J s–1 60 × 60 As
DQ DT = KA Dx Dt
Physics for you | november ‘15
29
\
Thermal conductivity of material DQ / Dt K= A(DT / Dx ) 233.33 × 4 × 10−3 or K = = 58.33 W m–1°C–1 −4 5 × 10 × 32
27. (c) : Let x kg of ice melts. Using law of calorimetry, heat lost by copper = heat gained by ice \ 2 × 400 × (500 – 0) = x × 3.5 × 105 or
x=
2 × 400 × 500
8 = kg 7
3.5 × 105 28. (a) : According to Wien’s displacement law lmaxT = constant
time : a t a rop d e p and n a O e h c , ifier r u A fast p r e t e wa effeacndtainv, Flexible AeplepctliroednicPhsysLaicbs,, IISC, bangalore
l max
T = 1 T2 l max 1
l max 4
2
=
2
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29. (c) 30. (c) : As
C F − 32 = 100 180
9 F = C + 32 5 Thus the graph between °C and °F is a straight line with positive intercept on y-axis as shown in the figure above. \
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Physics for you | november ‘15
ElEctrostatics Electrostatics is the branch of science that deals with the study of electric charges at rest. Here we study the forces, fields and potentials associated with static charges.
•
Electric charges
Charges are of two types, positive charge and negative charge. The charge developed on a glass rod when rubbed with silk is positive charge. The charge developed on a plastic rod when rubbed with wool is negative charge.
•
Basic properties of charge • •
•
•
•
•
Charge is a scalar quantity. Charge is transferable : If a charged body is put in contact with an uncharged body, the uncharged body becomes charged due to transfer of electrons from one body to the other. Charge is always associated with mass, i.e., charge can not exist without mass though mass can exist without charge. So, the presence of charge itself is a convincing proof of existence of mass. Quantization of charge : Total charge on a body is always an integral multiple of a basic unit of charge denoted by e and is given by q = ne where n is any integer, positive or negative and e = 1.6 × 10–19 C. The basic unit of charge is the charge that an electron or proton carries . By convention the charge on electron is –e (–1.6 × 10–19 C) and charge on proton is +e (1.6 × 10–19 C). Additivity of charge : Total charge of a system is the algebraic sum (i.e. sum is taking into account
•
with proper signs) of all individual charges in the system. Conservation of charge : Total charge of an isolated system remains unchanged with time. In other words, charge can neither be created nor be destroyed. Conservation of charge is found to hold good in all types of reactions either chemical or nuclear. Charge is invariant : Charge is independent of the frame of reference. Like charges repel each other while unlike charges attract each other.
charging of a Body
Charging a body means transfer of charges (electrons) from one body to the other. A body can be positively charged by losing some of its electrons and it can be negatively charged by gaining electrons. Methods of charging : A body can be charged by • Friction • Induction • Conduction Charging by induction is preferred because one charged body can be used to charge any number of uncharged bodies without any loss of charge. If q be the source of charge, then charge induced on a body of dielectric 1 constant K is given by q′ = − q 1 − K For metals, K = ∞ \ q′ = –q Physics for you | november ‘15
31
i.e., charges induced are equal and opposite only in case of conductors. In general, magnitude of induced charge is less than that of inducing charge. coulomb’s law
It states that, the electrostatic force between two stationary charges is proportional to the product of magnitude of charges and inversely proportional to the square of the distance between them. qq i.e., F ∝ 1 2 2 r qq 1 . 1 2 F= 4 pe0 r 2 1 = k = 9 × 109 N m 2C −2 Here 4pe 0 is proportionality constant. e0 = 8.854 × 10–12 C 2 N–1 m–2 is permittivity of free space. Vectorially Coulomb’s law can be written as q1q2 1 F12 = (r − r ) 4pe 0 | r1 − r2 |3 1 2 (The force on charge q1 due to charge q2) q1q2 1 (r − r ) and F21 = 4pe 0 | r1 − r2 |3 2 1 (The force on charge q2 due to charge q1) where the position vectors of charges q1 and q2 are r1 and r2. The Coulomb’s force between two charged particles in a medium is 1 q1q2 1 q1q2 Fmedium = = 4pe r 2 4pe 0e r r 2 F Fmedium = vacuum er KEY POINT • Permittivity is a measure of how an electric field
affects and is affected by a medium.
comparison between coulomb gravitational force are as follows : • • •
•
32
force
and
Coulomb force and gravitational force follow the same inverse square law. Coulomb force can be attractive or repulsive while gravitational force is always attractive. Coulomb force between the two charges depends on the medium between two charges while gravitational force is independent of the medium between the two bodies. The ratio of coulomb force to the gravitational force between two protons at same distance apart is e2 = 1.3 × 1036 4pe 0Gm pm p Physics for you | november ‘15
Superposition theorem : The interaction between any two charges is independent of the presence of all other charges. Electrical force is a vector quantity therefore, the net force on any one charge is the vector sum of all the forces exerted on it due to each of the other charges interacting with it independently i.e, Total force on charge q, F = F1 + F2 + F3 + ....... continuous charge Distribution
Linear charge density : Charge per unit length is known as linear charge density. It is denoted by symbol l. Charge l= Length Its SI unit is C m–1. Surface charge density : Charge per unit area is known as surface charge density. It is denoted by symbol s. Charge s= Area Its SI unit is C m–2. Volume charge density : Charge per unit volume is known as volume charge density. It is denoted by symbol r. Charge r= Volume Its SI unit is C m–3.
SELF CHECK
1. Two charges, each equal to q, are kept at x = – a and x = a on the x-axis. A particle of mass m and charge q q0 = is placed at the origin. If charge q0 is given a 2 small displacement (y < < a) along the y-axis, the net force acting on the particle is proportional to 1 (a) − (b) y y 1 (c) –y (d) y (JEE Main 2013) 2. Shown in the figure are two point charges +Q and –Q inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If s1 is the surface charge on the inner surface and Q1 net charge on it and s2 the surface charge on the outer surface and Q2 net charge on it then
charge +q0 at a point P distance r away, in a medium of permittivity e, can be calculated by imagining a very small charge +q to be placed at P.
(a) (b) (c) (d)
s1 ≠ 0, Q1 ≠ 0 ; s2 ≠ 0, Q2 ≠ 0 s1 ≠ 0, Q1 = 0 ; s2 ≠ 0, Q2 = 0 s1 ≠ 0, Q1 = 0 ; s2 = 0, Q2 = 0 s1 = 0, Q1 = 0 ; s2 = 0, Q2 = 0
(JEE Main 2015)
ElEctric fiElD
An electric field is a region where an electric charge experiences a force. If a very small, positive point charge Q is placed at any point in an electric field and it experiences a force F, then the electric field at that point is defined as E=
F Q
The magnitude of E is the force per unit charge and its direction is that of F(i.e., of the force which acts on a positive charge). Thus, electric field is a vector quantity. If F is in newton (N) and Q is in coulomb (C) then the unit of E is newton per coulomb (N C–1).
By Coulomb’s law, the force F on q is 1 qq0 , F ∴ E = 1 q0 E= F= 4 pe r 2 q 4pe r 2 as shown. If a point charge E is directed away from +q 0 –q0 is replaced by +q0, E would be directed towards –q0 since unlike charges attract. Electric field lines
An electric field can be represented and so visualized by electric field lines. These are drawn so that, the field lines at a point, (or the tangent to it if it is curved) gives the direction of E at that point, i.e., the direction in which positive charge would move and the number of lines per unit cross-section area is proportional to E. The field lines are imaginary but the field it represents is real. +q
–q
(i)
(ii)
Electric field due to a Point charge
The magnitude of E due to an isolated positive point
The friction caused by rapidly rising and falling currents of moving air creates electrical charges within a cloud. Water droplets and ice pellets fall, carrying charged electrons to the lower portion of the cloud, where a negative charge builds. A positive charge builds up near the top of a cloud. most of the electrical energy in a thunderstorm is dissipated within the clouds, as lightning hops between the positively and negatively charged areas. Lightning becomes dangerous, though, when it reaches earth. When the negative charge in the cloud becomes great enough, it seeks an easy path to the positively charged ground below. The current looks for a good conductor of electricity, or a tall structure anchored to the ground. As negative charges collect at the bottom of a storm cloud, a change happens on the ground below. electrons on the ground feel the power of the cloud’s negative charges. The electrons are pushed away from the area underneath the cloud. The ground and the objects on it are left with a positive charge. If you were standing on the ground below a storm cloud, you wouldn’t be able to see electrons move. but you might feel your skin tingle or your hair stand on end. As the ground becomes positively charged, the attraction between the cloud and the ground grows stronger. Suddenly, electrons shoot down from the cloud. They move in a path that reaches out in different directions—like the branches of a tree. each branch, or step, is about 45 m long. This branching path is called a stepped leader. After the first electrons have blasted their way through the air, other electrons from the cloud follow and make new branches. A stepped leader cuts through the air very quickly. Its average speed is about 1.2 × 105 m s–1. As the stepped leader nears the ground, a positive streamer reaches up for it. only then, once this channel is made, does the visible lightning happen. A return stroke runs from the ground to the clouds in a spectacular flash. Though the bolt appears continuous, it is actually a series of short bursts. most lightning strikes occur in less than a half second and the bolt is usually less than 5 cm in diameter.
Physics for you | november ‘15
33
The electric field due to a positive point charge is represented by straight lines originating from the charge as shown in figure (i). The electric field due to a negative point charge is represented by straight lines terminating at the charge as shown in figure (ii). The lines of force for a charge distribution containing more than one charge, is such that from each charge we can draw the lines isotropically. The lines may not be straight as one moves away from a charge. The shape of lines for some charge distribution is shown below.
φ= ∫ E ⋅ dS where ∫ represents closed integral done for a closed surface. The SI unit of electric flux is N m2 C–1 and its dimensional formula is [ML3T–3A–1]. ElEctric DiPolE
It is a pair of two equal and opposite charges separated by a small distance. Electric Dipole Moment
It is a vector quantity whose magnitude is equal to product of the magnitude of either charge and distance between the charges. i.e. | p | = q2a .
By convention the direction of dipole moment is from negative charge to positive charge. The SI unit of electric dipole moment is C m and its dimensional formula is [M0LAT]. The practical unit of electrical dipole moment is debye. Electric field intensity on axial line (End on Position) of the Electric Dipole
The lines of force are purely a geometrical construction which help us to visualise the nature of electric field in the region. They have no physical existence. The number of lines originating or terminating on a charge is proportional to the magnitude of charge. In rationalised MKS system (1/e0) electric lines are associated with unit charge. So if a body encloses a charge q, total lines of force associated with it (called flux) will be q/e0. Lines of force per unit area normal to the area of a point represents magnitude of intensity, crowded lines represent strong field while distant lines represent weak field. Electric flux
If the lines of force pass through a surface then the surface is said to have flux linked with it. It is given by dφ = E ⋅ dS where dS is the area vector of the small area element. The area vector of a closed surface is always in the direction of outward drawn normal. The total flux linked with whole of the body 34
Physics for you | november ‘15
At the point at a distance r from the centre of the electric 1 2 pr dipole, E = . 4pe 0 (r 2 − a 2)2 2p 4pe 0r 3 The direction of the electric field on axial line of the electric dipole is along the direction of the dipole moment (i.e. from –q to q). At very large distance i.e., (r > > a), E =
Electric field intensity on Equatorial line (broad on Position) of Electric Dipole
At the point at a distance r from the centre of electric 1 p dipole, E = . 4pe 0 (r 2 + a 2)3/2 At very large distance i.e., r > > a, 1 p E= . 4pe 0 r 3 The direction of the electric field on equatorial line of the electric dipole is opposite to the direction of the dipole moment (i.e. from q to –q).
Electric field intensity at any Point due to an Electric Dipole
The electric field intensity at point P due to an electric dipole, 1 p E= 1 + 3 cos 2 q 4pe 0 r 3 Electric field intensity due to a charged ring
At the centre of the ring, i.e. r = 0, E = 0. torque on an Electric Dipole Placed in a uniform Electric field
When an electric dipole of dipole moment p is placed in a uniform electric field E , it will experience a torque and is given by t = p×E or t = pE sin q
where q is the angle between p and E Torque acting on a dipole is maximum (tmax = pE) when dipole is perpendicular to the field and minimum (t = 0) when dipole is parallel or antiparallel to the field.
At a point on its axis at distance r from its centre, KEY POINT 1 qr E= • When a dipole is placed in a uniform electric field, 4pe 0 (r 2 + a 2)3/2 it will experience only torque and the net force on where q is the charge on the ring and a is the radius of the dipole will be zero while when it is placed in a the ring. non uniform electric field, it will experience both 1 q At very large distance i.e. r >> a, E = 2 torque and net force. 4pe 0 r List of formula for electric field intensity due to various types of charge distribution : Name /Type
Formula kq ^ E = 2 ⋅r |r |
Note f f
f
E=
l ^ 2kl r^ r= 2pe 0r r
f
f
f
s ^ E= n 2e 0
f
f
E=
kQx 2 (R + x 2)3/2
Ecentre = 0
f f
f
Graph
q is a source charge. r is vector drawn from source charge to the test point. outwards due to +ve charges and inwards due to –ve charges l is linear charge density (assumed uniform). r is perpendicular distance of point from line charge. r^ is radial unit vector drawn from the charge to test point. s is surface charge density (assumed uniform). ^
n is normal unit vector
E r
E /20 r
Q is total charge on the ring x = distance of point on the axis from centre of the ring. Electric field is always along the axis.
E Emax R
Physics for you | november ‘15
x
2 35
s^ E= n e0
f
f
(i) For r ≥ R, kQ ^ E= 2r |r |
f f
f
(ii) For r < R, E=0 f
(i) For r ≥ R, kQ ^ E= 2r |r | (ii) For r < R, kQ E= 3r R
^
R is radius of the sphere. r is vector drawn from centre of sphere to the point.
f
(s = surface charge density) r is vector drawn from centre of sphere to the point.
f
f f
Sphere acts like a point charge placed at the centre for points outside the sphere. E is always along radial direction. 4 3 Q is total charge r 3 pR (r = volume charge density) Inside the sphere, E ∝ r 1 Outside the sphere, E ∝ 2 r
SELF CHECK
3. A long cylindrical shell carries positive surface charge s in the upper half and negative surface charge –s in the lower half. The electric field lines around the cylinder will look like figure given in (Figures are schematic and not drawn to scale) (a)
(c)
+ + ++ + +
+ + ++ + +
(b)
+ + ++ + +
(d)
(JEE Main 2015) 36
Physics for you | november ‘15
r
E kQ/R2
Sphere acts like a point charge placed at centre for points outside the sphere. E is always along radial direction. Q is the total charge (= s4pR2)
f
/0
n is the unit vector perpendicular to the surface.
f
f
E
s is the surface charge density (assumed uniform).
R
r
E
kQ/R
2
R
r
4. A thin disc of radius b = 2a has a concentric hole of radius a in it (see figure). It carries uniform surface charge s on it. If the electric field on its axis at height h (h < < a) from its centre is given as Ch then value of C is (a)
s ae 0
(b)
s 2ae 0
(c)
s 4ae 0
(d)
s 8ae 0 (JEE Main 2015)
5. A wire, of length L(= 20 cm), is bent into a semicircular arc. If the two equal halves of the arc, were each to be uniformly charged with charges ±Q, [|Q| = 103 e0 Coulomb where e0 is the permittivity (in SI units) of free space] the net electric field at the centre O of the semi-circular arc would be
The total electric flux through a closed surface is zero if no charge is enclosed by the surface. In the situation when the surface is so chosen that there are some charges inside and some outside, the electric field (whose flux is calculated) is due to all the charges, both inside and outside the surface. However, the term (q) represents only the total charge inside the closed surface. KEY POINT • Gauss’s law is true for any closed surface, regardless
of its shape or size.
^
(b) (25 × 103 N/C) ^i
(a) (50 × 103 N/C) j 3
• The surface that we choose for the application of
^
3
(c) (25 × 10 N/C) j
Gauss’s law is known as the Gaussian surface.
^
• Gauss’s law is based on inverse square dependence
(d) (50 × 10 N/C) i
on distance.
(JEE Main 2015)
ElEctric PotEntial
Potential Energy of an Electric Dipole in a uniform Electric field is
Electric potential at a point is defined as amount of work done in bringing a unit positive charge from infinity to that point. It is denoted by symbol V. W V= q Electric potential is a scalar quantity. The SI unit of potential is volt and its dimensional formula is [ML2T–3A–1].
U = –pE(cosq2 – cosq1)
where q1 is the initial angle between p and E and q2 is the final angle between p and E . Gauss’s law
It states that the total electric flux through a closed surface S is 1/e0 times the total charge enclosed by S. q ∫ E ⋅ dS = e 0 where q is charge enclosed by the closed surface S.
Electric potential due a point charge q at a distance r from the charge is q V= 4pe 0r
Electric potential due to various charge distributions are given in table below : Name /Type Point charge
Ring (uniform/ non-uniform charge distribution)
Formula V=
kq r
kQ at centre , R kQ V= , along the axis R2 + x 2
V=
Uniformly charged kQ hollow conducting/ For r ≥ R , V = r non-conducting/solid For r ≤ R, V = kQ conducting sphere R
Note f f
f f
f f
f
Graph
q is source charge. r is the distance of the point from the point charge.
Q is charge on the ring. x is the distance of the point on the axis.
R is radius of sphere r is the distance from centre of sphere to the point Q is total charge = s4pR2
Physics for you | november ‘15
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Uniformly charged kQ solid non-conducting For r ≥ R, V = R sphere kQ(3R 2 − r 2) For r ≤ R, V = 2R 3 r = (3R 2 − r 2) 6e 0
f f
f f f f
Infinite line charge
Not defined
f
f
Infinite nonconducting thin sheet
Not defined
Infinite charged conducting thin sheet
Not defined
f
f
f f
SELF CHECK
6. A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ∞) on its surface. For this sphere the equipotential surfaces 3V 5V 3V V with potentials 0 , 0 , 0 and 0 have radius 2 4 4 4 R1, R2, R3 and R4 respectively. Then (a) (b) (c) (d)
R1 = 0 and R2 < (R4 – R3) 2R < R4 R1 = 0 and R2 > (R4 – R3) R1 ≠ 0 and (R2 – R1) > (R4 – R3)
(JEE Main 2015) Equipotential surface
An equipotential surface is a surface with a constant value of potential at all points on the surface. Properties of an equipotential surface Electric field lines are always perpendicular to an equipotential surface. Work done in moving an electric 38
Physics for you | november ‘15
R is radius of sphere r is distance from centre to the point 3 Vcentre = Vsurface 2 4 Q is total charge = r pR 3 3 Inside sphere potential varies parabolically Outside potential varies hyperbolically. Absolute potential is not defined. Potential difference between two points is given by formula VB – VA =– 2kl ln (rB/rA) Absolute potential is not defined Potential difference between two points is given by formula s VB − VA = − (r − r ) 2e 0 B A Absolute potential is not defined Potential difference between two points is given by formula s VB − VA = − (rB − rA) e0 charge from one point to another on an equipotential surface is zero. Two equipotential surfaces can never intersect one another. Relationship between E and V E = −∇V where ∇ = i^ ∂ + ^j ∂ + k^ ∂ ∂x ∂z ∂y –ve sign shows that the direction of E is the direction of decreasing potential. Electric Potential Energy
Electric potential energy of a system of charges is the total amount of work done in bringing the various charges to their respective positions from infinitely large mutual separations. The SI unit of electrical potential energy is joule. Electric potential energy of a system of two charges is 1 q1q2 U= 4pe 0 r12
where r12 is the distance between q1 and q2. Electric potential energy of a system of n point charges q jqk 1 U= r 4pe 0 all∑ pairs jk Note in this summation, we should include only one term for each pair of charges. conductors and insulators
On the basis of conductivity, all bodies can be divided in two classes, conductors and insulators. In conductors, electric charges are free to move throughout the volume. Insulators do not have free charges to move. Basic electrostatics properties of a conductor • Inside a conductor, electric field is zero. • At the surface of a charged conductor, electric field must be normal to the surface at every point. • The interior of a conductor can have no excess charge in the static situation. • Electric potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface. • Electric field at the surface of a charged conductor s E = n^ e0 where s is the surface charge density and n^ is a unit vector normal to the surface in the outward direction. Electrostatic shielding : It is the phenomenon of protecting a certain region of space from external electric field. Polar and non-Polar Molecules Dielectrics : Dielectrics are non conducting substances. In contrast to conductors, they have no (or negligible number of) charge carriers. Polar molecule : A polar molecule is one in which the centres of positive and negative charges are separated (even when there is no external field). A polar molecule has a permanent dipole moment e.g., water (H2O) and HCl. Non-polar molecule : A non-polar molecule is one in which the centres of positive and negative charges coincide. A non-polar molecule has no permanent dipole moment. e.g., oxygen (O2) and hydrogen (H2). Polarisation : The dipole moment per unit volume is called polarisation and is denoted by P. For linear isotropic dielectrics P = c e E , where ce is a constant characteristic of the dielectric and is called the electric susceptibility of the dielectric medium.
capacitor
A capacitor is a device that stores electrical energy. It consists of conductors of any shape and size carrying charges of equal magnitudes and opposite signs and separated by an insulating medium. Capacity of capacitor (Capacitance) Capacitance (C) of a capacitor is the ratio of charge(Q) given and the potential (V) to which it is raised. C=
Q V
The SI unit of capacitance is farad (F). 1 millifarad (mF) = 10–3 farad 1 microfarad (mF) = 10–6 farad 1 picofarad (pF) = 10–12 farad. Capacitance is a scalar quantity. The dimensional formula of capacitance is [M–1L–2T4A2]. Capacitance of spherical conductor Capacitance of a spherical conductor of radius R is C = 4pe0R Taking earth to be a conducting sphere of radius 6400 km, its capacitance will be 6.4 × 106 C = 4pe 0R = = 711 m F 9 × 109 Capacity of capacitor depends upon • Total outer surface area • Dielectric constant of the medium around the capacitor. • Presence of another capacitor in the near about region. Capacity of capacitor does not depend upon • Charge • Potential • Shape of capacitor • Material of capacitor types of capacitors
Parallel plate capacitor : C = e 0 A d (when air is between the plates) e KA (when dielectric is between the plates) C= 0 d Here, A is area of each plate and d is separation between the two plates. ab b−a Here, a and b are the radii of inner and outer coatings (spherical shells) of the spherical capacitor. Spherical capacitor : C = 4pe 0
Physics for you | november ‘15
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Cylindrical capacitor : l l C = 2pe 0 = 2pe 0 b b log e 2.303 log10 a a Here, a and b are the radii of the inner and outer coatings (cylindrical shells) and l is the length of the either curved surface of the cylindrical capacitor. combination of capacitors
Capacitors in series q
C1 V1
C2
C3 V2
q
V3
Same charge flows through each capacitor. Different potential difference exists across each capacitor. V = V1 + V2 + V3 1 1 1 1 = + + C C1 C2 C3 The resultant capacity C is less than the smallest capacitor joined in series. The ultimate effect is to reduce the capacitance in a circuit. Capacitors in parallel
Across each capacitor, potential difference is same. Different charges flow through each capacitor. q = q1 + q2 + q3 C = C1 + C2 + C3 The resultant capacity C is greater than the greatest capacitance joined in parallel. Thus the ultimate effect is to enhance the capacitance in a circuit. Energy stored in a capacitor
Work done in charging a capacitor gets stored in the capacitor in the form of its electrostatic potential energy. 1 1 1 Q2 U = CV 2 = QV = 2 2 2 C Electric Energy Density (uE )
The energy stored per unit volume in the electric field between the plates is called energy density. 1 s2 u E = e 0E 2 = . 2 2e 0 40
Physics for you | november ‘15
sharing of charges
When two capacitors charged to different potentials are connected by a conducting wire, charge flows from the one at higher potential to the other at lower potential till their potentials become equal. This equal potential is called common potential (V), where Total charge Q1 + Q2 C1V1 + C2V2 V= = = Total capacity C1 + C2 C1 + C2 It should be clearly understood that in sharing charges, there is absolutely no loss of charge. Some energy is, however, lost in the process which is given by C C (V − V )2 U1 − U 2 = 1 2 1 2 2(C1 + C2) This energy is potential energy. Effect of dielectric
When a dielectric slab of dielectric constant K is introduced between the plates of a charged parallel plate capacitor and the charging battery remains connected, then, potential difference between the plates remains constant i.e., V = V0 Capacitance C increases i.e., C = KC0 Charge on a capacitor increases i.e., Q = KQ0 Electric field between the plates remains unchanged i.e., E = E0 Energy stored in a capacitor increases i.e., U = KU0 When a dielectric slab of dielectric constant K is introduced between the plates of a charged parallel plate capacitor and the charging battery is disconnected, then, charge remains unchanged i.e., Q = Q0 Capacitance increases i.e., C = KC0 Potential difference between the plates decreases i.e., V V= 0 K Electric field between the plates decreases E i.e., E = 0 K Energy stored in the capacitor decreases i.e., U = U 0 K where Q0 , C0 , V0 , E0 and U0 represents the charge, capacitance, potential difference, electric field and energy stored in the capacitor of a charged air filled parallel plate capacitor. Van de Graaff generator A Van de Graaff generator consists of a large spherical conducting shell (a few metres in diameter). By
means of a moving belt and suitable brushes, charge is continuously transferred to the shell, and potential difference of the order of several million volts is built up, which can be used for accelerating charged particles.
SELF CHECK
7. In the given circuit, charge Q2 on the 2 mF capacitor changes as C is varied from 1 mF to 3 mF. Q2 as a function of ‘C’ is given properly by (figures are drawn schematically and are not to scale) 1 F
C
2 F E
Charge
Charge
(a) Q2
(b) 1 µF 3 µF
Q2
C
1 µF 3 µF
Charge
(c) Q2 1 µF 3 µF
C
Charge C
(d) Q2 1 µF
3 µF
C
(JEE Main 2015) 8. In figure is shown a system of four capacitors connected across a 10 V battery. Charge that will flow from switch S when it is closed is
The unit of current is ampere (A). 1A =1
C s
The dimentional formula of current is [M°L°T°A] Drift Velocity
It is defined as the average velocity with which free electrons get drifted towards the positive end of the conductor under the influence of an external electric field. Drift velocity of electrons is given by eE vd = − t m where –e is the charge and m is the mass of an electron, E is the electric field applied and t is known as relaxation time. The value of drift velocity of an electron is about 10–4 m s–1 and value of relaxation time is about 10–14 second. The direction of drift velocity of electrons in a metal conductor is opposite to that of electric field applied E. Drift velocity depends on electric field as vd ∝ E. So greater the electric field, larger will be the drift velocity. The thermal speed or rms speed of electrons at room temperature is about 105 m s–1, which is very large as compared to the drift velocity of electrons. mean free path Relaxation time = rms speed of electrons relationship between current and drift velocity
I = nAevd where n is the number density of electrons or number of electrons per unit volume of the conductor and A is the area of cross-section of the conductor. Mobility
(a) 5 mC from b to a (c) 5 mC from a to b
(b) 20 mC from a to b (d) zero
(JEE Main 2015)
currEnt ElEctricity Electric current
It is the amount of charge flowing across any section of wire per unit time. If the moving charges are positive, the current is in the direction of motion of charges. If they are negative the current is opposite to the direction of motion of charges. Instantaneous current is the current at any point of time dq q = and average current = . dt t
It is defined as the magnitude of drift velocity per unit electric field. It is denoted by symbol m. v qEt / m qt m= d = = E E m where q, t and m are charge, relaxation time and mass of a charge carrier respectively. Mobility is positive for both electrons and holes although their drift velocities are opposite to each other. The SI unit of mobility is m2 V–1 s–1 and its dimensional formula is [M–1L0T2A]. ohm’s law
It was discovered by German physicist Georg Simon Ohm in 1828. It states that the current (I) flowing through a conductor is directly proportional to the potential difference (V) across the ends of the conductor, Physics for you | november ‘15
41
provided physical conditions of the conductor such as temperature, mechanical strain etc. are kept constant. V ∝ I or V = RI where the constant of proportionality R is called resistance of the conductor. V The graph between potential difference (V) and current (I) through a metallic conductor is slope = tan a straight line passing through I the origin as shown in figure. O The slope of V-I graph gives resistance. V R = = tan q (slope of V -I graph) I Electrical resistance The resistance of a conductor is the obstruction posed by the conductor to the flow of current through it. Resistance of a conductor (R) is defined as the ratio of potential difference (V) applied across the ends of conductor to the current (I) flowing through it. V R= I Resistance is a property of the conductor and is not related to the circuit in which the conductor is connected. The V equation R = can be used to calculate the current in I a conductor if we know the resistance and the potential difference across the conductor. The SI unit of resistance is ohm (W) and its dimensional formula is [ML2T–3A–2]. The resistance of a conductor not only depends on the material of the conductor but also on the dimensions of the conductor. The resistance of a conductor is proportional to its length and inversely proportional to its area of crosssection. l l R∝ or R = r A A where constant of proportionality r is called resistivity. The SI unit of resistivity is W m and its dimensional formula is [ML3T–3A–2]. m The resistivity of a conductor is r = 2 ne t where m is the mass and e is the magnitude of charge of an electron, n is the number density of electrons, t is the relaxation time. The resistivity of a conductor is independent of its dimensions but depends on the material of the conductor. A perfect conductor would have zero resistivity and a perfect insulator would have infinite resistivity. 42
Physics for you | november ‘15
If the conductor is in the form of wire of length l and rl radius r, then its resistance is R = 2 . pr If a conductor has mass m, volume V and density d, then its resistance R is rl rl 2 rl 2 rl 2d R= = = = A Al V m If a wire of resistance R is cut into n equal parts, then resistance of each part = R/n. If length of a given metallic wire of resistance R is stretched to n times, its resistance becomes n2R. If radius of the given metallic wire of resistance R becomes n times, its resistance becomes (1/n4)R. If the area of cross-section of the given metallic wire of resistance R becomes n times, then its resistance becomes (1/n2)R. A cylindrical tube of length l has inner and outer radii r1 and r2 respectively. The resistance between its end faces rl is R = . 2 p r2 − r12 If a rectangular conducting cube having dimensions l × b × h, then • when current passes through the length side, the rl resistance offered by cube = bh • when current passes through breadth side, the rb resistance offered by cube = lh
(
)
when current passes through the thickness side, the rh resistance offered by cube = . lb current density, conductance and conductivity Current density at a point inside the conductor is defined as the amount of current flowing per unit area around that point of the conductor, provided the area is held in a direction normal to the current. It is denoted by symbol J. I J= A If area A is not normal to the current but makes an angle q with the direction of current, then I J= or I = JA cos q = J ⋅ A A cos q •
The SI unit of current density is A m–2 and its dimensional formula is [M0L–2T0A]. Current density is a vector quantity. Its direction is that of the flow of positive charge at the given point inside the conductor.
KEY POINT • Current density is a characteristic property of a
particular point inside the conductor and not of the whole conductor. • Current is the flux of current density.
relationship between current density and drift velocity
J = nqvd where n is the number density of charge carriers each of charge q, and vd is the drift velocity of the charge carriers. For electrons q = –e. Conductance : The reciprocal of resistance is called conductance. It is denoted by symbol G. 1 G= R The SI unit of conductance is ohm–1 which is called mho and is represented by the symbol ( ). The unit mho is also called siemen (S) and its dimensional formula is [M–1L–2T3A2]. Conductivity : The reciprocal of resistivity is called conductivity or specific conductance. It is denoted by symbol s. 1 ne 2t v d et s= = = nem As m = E = m r m The SI unit of conductivity is W–1 m–1 or S m–1 or mho m–1 and its dimensional formula is [M–1L–3T3A2]. relationship between J, s and E
J = sE It is a microscopic form of Ohm’s law. ohmic and non-ohmic conductors
Ohmic conductors : Those conductors which obey Ohm’s law are called ohmic conductors, e.g. metals. For Ohmic conductors, V-I graph is a straight line passing through the origin. Non-ohmic conductors : Those conductors which do not obey Ohm’s law are called non-ohmic conductors e.g. diode valve, junction diode. For non-ohmic conductors V-I graph is non-linear. Effect of temperature on resistance
The resistance of a metallic conductor increases with increase in temperature. The resistance of a conductor at temperature T°C is given by RT = R0 (1 + aT + bT2) where RT is the resistance at T°C, R0 is the resistance at 0°C and a and b are the characteristics constants of the material of the conductor. If the temperature T°C is not sufficiently large, b is negligible, the above relation can be written as
RT = R0 (1 + aT) where a is the temperature coefficient of resistance. Its unit is K–1 or °C–1. For metals, a is positive, therefore resistance of a metal increases with rise in temperature. For insulators and semiconductors, a is negative therefore their resistance decreases with rise in temperature. For alloys like manganin and constantan, the value of a is very small as compared to that for metals. Due to high resistivity and low temperature coefficient of resistance, these alloys are used in making standard resistance coils. thermistors
A thermistor is a heat sensitive device whose resistivity changes very rapidly with change of temperature. colour code for resistors
A colour code is used to indicate the resistance value and its percentage accuracy. Every resistor has a set of coloured rings on it. The first two coloured rings from the left end indicate the first two significant figures of the resistance in ohms. The third colour ring indicates the decimal multiplier and the last colour ring stands for the tolerance in percent. The colour code of a resistor is as shown in the table. Colour
Number
Multiplier
Tolerance (%)
Black 0 100 Brown 1 101 Red 2 102 Orange 3 103 Yellow 4 104 Green 5 105 Blue 6 106 Violet 7 107 Gray 8 108 White 9 109 Gold 10–1 5 –2 10 Silver 10 20 No colour Suppose a resistor has green, red, orange and gold rings as shown in the figure below. The resistance of the resistor is (52 × 103 W) ± 5%.
Physics for you | november ‘15
43
combination of resistors
SELF CHECK
resistors in series
The various resistors are said to be connected in series if they are connected as shown in the figure. R1
I
R2
I
I
V2
V1
R3 V3 I
I
10. Suppose the drift velocity vd in a material varied
–
+
I
9. When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10–4 m s–1. If the electron density in the wire is 8 × 1028 m–3, the resistivity of the material is close to (a) 1.6 × 10–6 W m (b) 1.6 × 10–5 W m –8 (c) 1.6 × 10 W m (d) 1.6 × 10–7 W m (JEE Main 2015)
The current through each resistor is the same. The equivalent resistance of the combination of resistors is Rs = R1 + R2 + R3 As current through each resistor is same so V1 V2 V3 = = R1 R2 R3 or V1 : V2 : V3 = R1 : R2 : R3 Series combinations of resistors are used in resistance box and decorative bulbs. resistors in parallel
with the applied electric field E as v d ∝ E . Then V-I graph for a wire made of such a material is best given by V
(a)
(b) I
I
V
(c)
(d) I
R1
R3
I3
I
(JEE Main 2015)
Electric cell
R2
I2
I +
–
The potential difference is same across each resistor (say V). The equivalent resistance of the combination of resistors is 1 1 1 1 = + + R p R1 R2 R3 V V V As I1 = , I 2 = and I 3 = so, R1 R2 R3 1 1 1 I1 : I 2 : I 3 = : : R1 R2 R3 All the domestic appliances in a house are connected in parallel combination. KEY POINT • If n wires each of resistance R are connected in
series, the equivalent resistance is nR. • If n wires each of resistance R are connected in parallel, the equivalent resistance is R/n. 44
I
V
The various resistors are said to be connected in parallel if they are connected as shown in figure below. I1
V
Physics for you | november ‘15
An electric cell is a device which maintains a continuous flow of charge in a circuit by a chemical reaction. Electromotive force (emf) of a cell
It is defined as the potential difference between the two terminals of a cell in an open circuit i.e., when no current flows through the cell. It is denoted by symbol e. The SI unit of emf is J C–1 or volt and its dimensional formula is [ML2T–3A–1]. The emf of a cell depends upon the nature of electrodes, nature and the concentration of electrolyte used in the cell and its temperature. terminal potential difference of a cell
It is defined as the potential difference between two terminals of a cell in a closed circuit i.e. when current is flowing through the cell. It is generally denoted by symbol V and is measured in volt. internal resistance of a cell
It is defined as the resistance offered by the electrolyte and electrodes of a cell when the current flows through it. Internal resistance of a cell depends upon the following factors:
• • • •
Distance between the electrodes The nature of the electrolyte The nature of electrodes Area of the electrodes, immersed in the electrolyte KEY POINT
n I
I R
Grouping of cells
Equivalent emf of the cells is eeq = ne Equivalent internal resistance of the cells is req = nr ne Current in the circuit, I = R + nr Special cases : ne e If R < < nr, then I = = . nr r ne If R > > nr, then I = . R KEY POINT • When one cell is wrongly connected in series of n identical cells, each of emf e, it will reduce the total emf by 2e i.e. effective emf = ne – 2e. • The total internal resistance of cells = nr i.e. there is no effect on the total internal resistance of the cells. • In series grouping of cells their emf ’s are additive or subtractive while their internal resistances are always additive. Parallel grouping If m identical cells each of emf e and internal m resistance r are connected to the external resistor of I I resistance R as shown in figure, they are said to R be connected in parallel grouping. Equivalent emf of the cells is eeq = e r Equivalent internal resistance of the cells is req = . m e Current in the circuit, I = . r R+ m Special cases : r e If > R, then I = m . m r
Cells can be grouped in the following three ways: Series grouping If n identical cells each of emf e and internal resistance r are connected to the external resistor of resistance R as shown in the figure, they are said to be connected in series grouping.
If the cells are connected as shown in figure they are said to be connected in mixed grouping. Let there be n cells in series in one row and m such rows of cells in parallel. Suppose all the cells are identical. Let each cell be of emf e and internal resistance r.
• A cell is a source of constant emf but not a constant
current. • The emf is not a force.
relationship between e, V and r
When a cell of emf e and internal resistance r is connected to an external resistance R as shown in the figure. r I
I R
Current in the circuit, I =
e R+r
Terminal potential difference, V = IR = Internal resistance of a cell, e −V e r= R = − 1 R V V
eR (R + r)
During discharging of a cell, terminal potential difference = emf of a cell – voltage drop across the internal resistance of a cell i.e. terminal potential difference across it is lesser than emf of the cell. The direction of current inside the cell is from negative terminal to positive terminal. During charging of a cell, terminal potential difference = emf of a cell + voltage drop across internal resistance of a cell i.e. terminal potential difference becomes greater than the emf of the cell. The direction of current inside the cell is from positive terminal to negative terminal. When the cell is short circuited, i.e., R = 0, the maximum current is drawn from a cell whose value is e given by I max = r
Mixed grouping
Physics for you | november ‘15
45
I1 – I2 – I3 – I4 + I5 = 0 ⇒ I1 + I5 = I2 + I3 + I4 This law is based on conservation of electrical charge.
n m
second law (Loop rule) I
I R
Equivalent emf of the cells is eeq = ne Equivalent internal resistance of the cells is nr m ne Current in the circuit, I = nr R+ m
req =
In case of mixed grouping of cells, current in the circuit will be maximum, when R=
nr m
i.e. external resistance = total internal resistance of all cells Potential Divider
If two resistors R1 and R2 are joined in series and a potential difference V is applied across them, and an output taken from the point P, at the joint of the two resistors, and the opposite end of R2, the arrangement is known as a potential divider. The potential difference across R2 is V2 where V2 = R2 V . R1 + R2
SELF CHECK
R1 V
P R2
Output (V2)
Kirchhoff’s Laws
There are two laws given by Kirchhoff for determination of potential difference and current in different branches of any complicated network. first law (Junction rule)
In an electric circuit, the algebraic sum of the current meeting at any junction is zero. I2 Sum of the currents entering I3 the junction is equal to sum I1 of the currents leaving I5 the junction. SI = 0 48
In any closed circuit, the algebraic sum of emfs and algebraic sum of potential drop is zero. SIR + Se = 0 This law is based on law of conservation of energy. While moving from negative to positive terminal inside the cell, emf is taken as positive while moving in the direction of current in a circuit the potential drop (i.e., IR) across resistance is taken as negative. When we move through a resistor in the same direction as current, the IR term is negative because the current goes in the direction of decreasing potential. When we move through a resistor in the direction opposite to the assumed current, the IR term is positive because this represents a rise of potential. According to sign convention while traversing a closed loop (in clockwise or anti-clockwise direction), if negative pole of the cell is encountered first then its emf is positive, otherwise negative. The product of resistance and current in an arm of the circuit is taken positive if the direction of current in that arm is in the same sense as one moves in a closed loop and is taken negative if the direction of current in that arm is opposite to the sense as one moves in the closed loop.
Physics for you | november ‘15
I4
11. In the circuit shown, the current in the 1 W resistor is (a) 0.13 A, from Q to P (b) 0.13 A, from P to Q (c) 0.3 A, from P to Q (d) 0 A
6V
P 2Ω 1Ω
9V
Q 3Ω
3Ω
(JEE Main 2015)
12. A 10 V battery with internal resistance 1 W and a 15 V battery with internal resistance 0.6 W are connected in parallel to a voltmeter (see figure). The reading in the voltmeter will be close to (a) 11.9 V (b) 12.5 V (c) 13.1 V (d) 24.5 V
10 V
1
15 V
0.6
V
(JEE Main 2015)
13. In the electric network shown, when no current flows through the 4 W resistor in the arm EB, the potential difference between the points A and D will be
(a) 3 V (c) 5 V
(b) 4 V (d) 6 V
where K is known as potential gradient i.e., fall of potential per unit length of the given wire. comparison of emfs of two cells by using potentiometer
(JEE Main 2015)
Wheatstone’s Bridge
It is an arrangement of four resistances P, Q, R and S connected as shown in the figure. Their values are so adjusted that the galvanometer G shows no deflection. The bridge is then said to be balanced. When this happens, the points B and D are at the same potential and no current flows through galvanometer and it can P R be shown that = Q S This is called the balancing condition. If any three resistances are known, the fourth can be found.
e1 l1 = e 2 l2
where l1, l2 are the balancing lengths of potentiometer wire for the emfs e1 and e2 of two cells respectively. Determination of internal resistance of a cell by potentiometer
Metre bridge or slide metre bridge
It is based on the principle of Wheatstone’s bridge.
l1 − l2 R Internal resistance of cell, r = l2 where l1 = balancing length of potentiometer wire corresponding to emf of the cell, l2 = balancing length of potentiometer wire corresponding to terminal potential difference of the cell when a resistance R is connected in series with the cell whose internal resistance is to be determined. Joule’s Law of heating
The unknown resistance, R = Sl 100 − l where l is the balancing length of metre bridge. Potentiometer
Principle of potentiometer : It is based on the fact that the fall of potential across any portion of the wire is directly proportional to the length of that portion provided the wire is of uniform area of cross-section and a constant current is flowing through it. i.e., V ∝ l (If I and A are constants) or V = Kl
According to Joule’s heating effect of current, the amount of heat produced (H) in a conductor of resistance R, carrying current I for time t is H = I2Rt (in joule) I 2Rt (in calorie) or H = J where J is Joule’s mechanical equivalent of heat (= 4.2 J/cal). Electric Power
It is defined as the rate at which work is done by the source of emf in maintaining the current in the electric circuit. Physics for you | november ‘15
49
Electric power, P =
Electric work done Time taken
V2 P = VI = I 2R = . R The SI unit of power is watt (W). The bigger unit of power is kilowatt (kW). 1 kilowatt = 1000 watt. The commercial unit of power is horse power (hp) 1 hp = 746 watt. Electric energy It is defined as the total electric work done or energy supplied by the source of emf in maintaining the current in an electric circuit for a given time. Electric energy = electric power × time V2 = Pt = VIt = I 2Rt = t R The SI unit of electrical energy is joule (J). The commercial unit of electric energy is kilowatt-hour (kWh), or Board Trade Unit (BTU). 1 kWh = 1000 Wh = 3.6 × 106 J Number of units of electricity consumed watt × hour = No. of kWh = 1000 Maximum power transfer theorem It states that the power output across load due to a cell or battery is maximum if the load resistance is equal to the effective internal resistance of cell or battery. fuse wire It is generally made up of tin-lead alloy. It should have high resistivity and low melting point. It is used in series of the electrical installations to protect them from strong currents.
50
Physics for you | november ‘15
If P1, P2, P3 .... are the powers of electric appliances in series with source of rated voltage V, the effective power consumed is 1 1 1 1 = + + + ...... PS P1 P2 P3 If P1, P2, P3.... are the powers of electric appliances in parallel with a source of rated voltage V, the effective power consumed is PP = P1 + P2 + P3 + .....
SELF CHECK
14. In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be (a) 14 A (b) 8 A (c) 10 A (d) 12 A (JEE Main 2014) 15. The supply voltage to a room is 120 V. The resistance of the lead wires is 6 W. A 60 W bulb is already switched on. What is the decrease of voltage across the bulb, when a 240 W heater is switched on in parallel to the bulb? (a) 10.04 V (b) zero V (c) 2.9 V (d) 13.3 V (JEE Main 2013) ansWEr KEys (sELf chEcK) 1. (b) 2. (c) 6. (a,b) 7. (d) 11. (a) 12. (c)
3. (c) 8. (a) 13. (c)
4. (c) 9. (b) 14. (d)
5. (b) 10. (c) 15. (a) nn
1. The plates of a parallel plate capacitor are charged up to 100 V. A 2 mm thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by 1.6 mm. The dielectric constant of the plate is (a) 5 (b) 1.25 (c) 4 (d) 2.5 2. In the circuit shown in figure, resistance of each resistor is R. The value of current I is 10V0 B (a) 7R R R R 15V0 R R (b) A C 7R E I R 7V0 R R (c) 8R V0 D 15V0 (d) 8R 3. One plate of a capacitor is connected to a spring as shown in figure. Area of both the plates is A. In steady state, separation between the plates is 0.8 d (spring was unstretched and the distance between the plates was d when the capacitor was uncharged). The force constant of the spring is approximately 6 ∈0 V 2 4 ∈0 AV 2 (a) (b) Ad 3 d3 (c)
∈0 AV 3
(d)
2 ∈0 AV
d2 2d 4. Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon, such that the electric field at O is double the electric field when only one positive charge of same magnitude P Q is placed at R, as shown in figure. Which of the U O R following arrangement of charges is possible for P, Q, S R, S, T and U respectively? T 3
(a) + – + – – + (c) + + – + – –
(b) + – + – + – (d) – + + – + –
5. A point charge q1 = 9.1 mC is held fixed at origin. A second point charge q2 = –0.42 mC and of mass 3.2 × 10–4 kg is placed on the x-axis 0.96 m from the origin. The second point charge is released at rest. What is its speed when it is 0.24 m from the origin? (a) 2.4 m s–1 (b) 9.6 m s–1 (c) 26 m s–1 (d) 48 m s–1 6. At room temperature, copper has free electron density of 8.4 × 1028 m–3. The electron drift velocity in a copper conductor of cross-sectional area of 10–6 m2 and carrying a current of 5.4 A, will be (a) 4 m s–1 (b) 0.4 m s–1 –1 (c) 4 cm s (d) 0.4 mm s–1 7. A battery of emf 2.1 V and internal resistance 0.05 W is shunted for 5 s by a wire of constant resistance 0.02 W, mass 1 g and specific heat 0.1 cal g–1°C–1. The rise in the temperature of the wire is (a) 10.7°C (b) 21.4°C (c) 107°C (d) 214°C 8. Two small spheres of masses M1 and M2 are suspended by weightless insulating threads of lengths L1 and L2. The spheres carry charges Q1 and Q2 respectively. The spheres are suspended such that they are in level with one another and the threads are inclined to the vertical at angles of q1 and q2 as shown. Which one of the following conditions is essential, if q1 = q2? (a) M1 ≠ M2, but Q1 = Q2 (b) M1 = M2 (c) Q1 = Q2 (d) L1 = L2 9. A solid conducting sphere of radius a having a charge q is surrounded by a concentric conducting spherical shell of inner radius 2a and outer radius 3a as shown in figure.
a
S 3a
Physics for you | November ‘15
2a
51
Find the amount of heat produced when switch is closed. 1 Here, k = 4 πe 0
kq2 (a) 2a (c)
(b)
kq2 4a
(d)
13. A conical wire of length l has end diameters a and b as shown in figure. The resistivity of the material of wire is r. The resistance of the wire between its two ends is
kq2 3a kq2 6a
a
b
10. Three infinitely long charge sheets are placed as shown in figure. The electric field at point P is
l
(a) (c)
4rl 2
(b)
2
π(a + b ) 4rl
(d)
π(a + b)2
4rl πab 16rl π(a + b)2
14. A conductor with rectangular cross-section has dimensions (a × 2a × 4a) as shown in figure. Resistance across AB is R1, across CD is R2 and across EF is R3. Then 2σ k e0 4σ k (c) e0
C
2σ k e0 4σ k (d) − e0
(b) −
(a)
A
11. If an electron enters into a space between the plates of a parallel plate capacitor at an angle a with the plates and leaves at an angle b to the plates, the ratio of its kinetic energy while entering the capacitor to that while leaving will be sin b (a) sin a
2
cos b (b) cos a
2
sin a (d) sin b
cos a (c) cos b
2
2
a
2a E
(a) R1 = R2 = R3 (c) R2 > R3 > R1
D
(b) R1 > R2 > R3 (d) R1 > R3 > R2
U
U
(a) –x
O
x (b) –x
U
52
R1 is 32 W R2 is 35 W total resistance of circuit is 48 W total resistance of circuit is 75 W Physics for you | November ‘15
B
4a
15. Four equal charges of magnitude q each are placed at four corners of a square with its centre at origin and lying in y-z plane. A fifth charge +Q is moved along x-axis. The electrostatic potential energy (U) varies on x-axis as shown in figure.
12. In the circuit shown in figure, the value of
(a) (b) (c) (d)
F
(c) –x
O
O
x
U
x (d) –x
O
x
16. A number of condensers, each of the capacitance 1 mF and each one of which gets punctured if a potential difference just exceeding 500 V is applied, are provided. An arrangement suitable for giving capacitance of 2 mF across which 3000 V may be applied requires at least (a) 6 component capacitors (b) 12 component capacitors (c) 72 component capacitors (d) 2 component capacitors 17. A charge q is placed at the centroid of an equiliateral triangle. Three charges equal to Q are placed at the vertices of the triangle. The system of four charges will be in equilibrium if q is equal to (a) −Q 3 (b) –Q/3 (c) −Q / 3
(d) Q/ 3
18. Consider a non-conducting shell as shown in figure. Two point charges are inside the shell and two are outside the shell. If we apply Gauss’s law over the non-conducting shell as Gaussian surface, the E on LHS of Gauss equation is due to q2 q (a) q1 and q2 alone 3 q1 (b) all charges q1, q2, q3 and q4 q4 (c) q1, q2 and q3 alone (d) We cannot take the non-conducting shell as a Gaussian surface 19. Three concentric spherical conductors are arranged as shown in the figure. The potential at point P will be 1 Q1 Q2 Q3 + + (a) r r 4 πe0 r (b)
1 Q1 + Q2 Q3 + c 4 πe0 r
Q3 Q2
Q1 C B A a
r
P
b c
Q1 Q2 Q3 a + b + c Q1 Q2 Q3 c + c + c 20. Two identical particles of charge q each are connected by a massless spring of force constant k. They are placed over a smooth horizontal surface. They are released when the separation between them is r and spring is unstretched. If maximum extension of the spring is r, the value of k is (neglect gravitational effect) 1 (c) 4 πe0 1 (d) 4 πe0
(a) (b) (c)
q2
q
8πe0r 3
k
q2
q
4πe0r 3 q2
(d)
πe0r 3
4q 2
πe0r 3
21. A ring of radius R is made out of a thin metallic wire of area of cross section A. The ring has a uniform charge Q distributed on it. A charge q0 is placed at the centre of the ring. If Y is the Young’s modulus of the material of the ring and DR is the change in the radius of the ring then q0Q qQ (a) DR = (b) DR = 2 0 4πe0 RAY 4π e RAY 0
q0Q
q0Q
(d) DR = 2 8π e0 RAY 8π2 e20 RAY 22. Uniform electric field exists in a region and is given by E = E0 i + E0 j . There are four points A(–a, 0), B(0, –a), C(a, 0) and D(0, a) lying in x – y plane. Which of the following is the correct relation for the electric potential ? (a) VA = VC > VB = VD (b) VA = VB > VC = VD (c) VA > VC > VB = VD (d) VA < VC < VB < VD (c) DR =
23. A circuit consists of a battery, a resistor R and two light bulbs A and B as shown in figure. If the filament in light bulb A burns out, then which of the following is true for light bulb B? (a) It is turned off. (b) Its brightness does not change. (c) It gets dimmer. (d) It gets brighter. 24. Charges +q and –q are placed at points A and B respectively which are at distance 2L apart, C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is qQ qQ (a) 2 πe L (b) 6 πe L 0 0 qQ qQ (c) − (d) − 4 πe0 L 6 πe0 L Physics for you | November ‘15
53
25. In the Wheatstone bridge shown in the figure, in order to balance the bridge, we must have
(a) R1 = 3 W; R2 = 3 W (b) R1 = 6 W; R2 = 15 W (c) R1 = 1.5 W; R2 = any finite value (d) R1 = 3 W; R2 = any finite value 26. A resistance of 4 W and a wire of length 5 m and resistance 5 W are joined in series and connected to a cell of emf 10 V and internal resistance 1 W. A parallel combination of two identical cells is balanced across 3 m of the wire. The emf E of each cell is (a) 1.5 V (b) 3.0 V (c) 0.67 V (d) 1.33 V 27. In the circuit shown, A and V are ideal ammeter and voltmeter respectively. Reading of the voltmeter will be 2V (a) 2V (b) 1 V A V (c) 0.5 V 1 1 (d) zero 28. In the figure, a carbon resistor has bands of different colours on it as shown in the figure. The value of the resistance is Red
Silver
White Brown
(a) (2.2 ± 10%) kW (c) (5.6 ± 5%) kW
(b) (3.1 ± 5%) kW (d) (9.1 ± 10%) kW
29. At room temperature (27°C) the resistance of a heating element is 100 W. What is the temperature of the element if the resistance is found to be 117 W given that the temperature coefficient of the material of the resistor is 1.70 × 10–4 °C–1 ? (a) 300°C (b) 627°C (c) 1027°C (d) 1700°C 54
Physics for you | November ‘15
30. Rated power of a bulb at voltage V is P. Now same voltage V is applied in all conditions mentioned in column I. Match this column I with column II in which actual total power consumed is given. Column I P. Two bulbs are connected in parallel. Q. Two bulbs are connected in series. R. Two bulbs are conncected in parallel and one bulb in series with this combination. S. A group of two-two bulbs in parallel are mutually connected in series. (a) P – 2, Q – 3, R – 1, S – 4 (b) P – 3, Q – 2, R – 4, S – 1 (c) P – 2, Q – 3, R – 4, S – 1 (d) P – 3, Q – 2, R – 1, S – 4
Column II 1. P 2.
2P
3.
P 2
4.
2P 3
sOLUTIONs 1. (a) : Here, t = 2 mm, x = 1.6 mm As potential difference remains the same, capacity must remain the same. 1 \ x = t 1 − K 1 1.6 = 2 1 − , which gives K = 5 K 2. (b) : In the given circuit, potential at B = potential at D. Therefore, points B and D can be joined together. Due to it, the arm AB will be parallel to arm AD. The arm BE will be parallel to arm ED and arm BC will be parallel to arm CD. The equivalent circuit will be as shown in figure. The resistance of arm BE is parallel to resistance of arm BCE. The effective resistance between B and E R R + R 3R = = 2 2 R R 8 + + R 2 2
Resistance of arm ABE R 3R 7 R = + = 2 8 8 7R V0 + R 8 15V0 \ Current, I = = 7R 7R ×R 8 3. (b) : In equilibrium, electrostatic force of attraction between the plates = restoring force in the string q2 = kx 2e0 A
7. (d) : Given, e = 2.1 V, r = 0.05 W, t = 5 s, R = 0.02 W, m = 1 g and s = 0.1 cal g–1°C–1 As per question,
or
(CV )2 = k(d − 0.8d ) = 0.2kd 2e 0 A
C 2V 2 2e0 A(0.2d ) e A Now C = 0 0.8d \
k=
\
k=
DT = 214°C 8. (b) :
e20 A2
4e0 AV 2 V2 = 0.64d 2 2e0 A(0.2d ) d3
4. (d) : According to option (d), electric fields at O, due to P and S will cancel. Again electric fields at O due to Q and T will cancel. Electric fields at O due to U and R will add up to give double the electric field. 5. (c) : From conservation of mechanical energy, we have decrease in gravitational potential energy = increase in kinetic energy. or
I 2 Rt 4. 2 2 e Rt m × s × DT = R + r 4. 2 2.1 × 2.1 0.02 × 5 1 × 0.1 × DT = × 0.07 × 0.07 4. 2 90 900 DT = = 0 . 1 × 4 . 2 4. 2 m × s × DT =
q q 1 1 1 2 mv = U i − U f = 1 2 − 2 4 πe0 ri r f
For sphere 1, in equilibrium T1 cos q1 = M1g and T2 sin q1 = F1 F \ tan q1 = 1 M1 g
F2 M2 g F is same on both the charges, so q will be same only if their masses are equal. 9. (c) : Similarly for sphere 2, tan q2 =
q q r f − ri = 1 2 4πe0 ri r f or =
v=
q1q2 r f − ri 2πe0m ri r f
(9.1 × 10−6 )(−0.42 × 10−6 ) × 2 × 9 × 109 0.24 − 0.96 (0.24)(0.96) 3.2 × 10−4
= 26 m s–1
6. (d) : Drift velocity in a copper conductor I 5. 4 v= = 28 neA 8.4 × 10 × 1.6 × 10−19 × 10−6 = 0.4 × 10–3 m s–1 = 0.4 mm s–1
Heat produced = Ui – Uf = (U1 + U2) – U2 = U1 =
kq2 q2 q2 = = 2C a × 2a 4a 2 4 πe0 a − a 2
{ }
10. (b) : Using principle of superposition of electric field, the total electric field intensity at P due to various plane sheets of charge will be Physics for you | November ‘15
55
σ 2σ σ EP = (−k ) + (− k ) + (−k ) 2e 0 2e 0 2e 0 2σ = − k e0
11. (b) : Suppose, u = velocity of electron while entering the field v = velocity of electron while leaving the field As electric field is perpendicular to the plates, component of velocity parallel to plates will remain unchanged. \ u cos a = v cos b 2 Ke u2 cos b u cos b = \ = = or v cos a K l v 2 cos a 12. (c) : Potential difference across C and D = 0.05 × 120 = 6 V Potential difference across B and E = Potential difference across C and D = 6 V 6 \ R2 = = 30 W 0. 2 Potential difference across R1 = 12 – 6 = 6 V = (0.05 + 0.2) R1 6 = 24 W or R1 = 0.25 Total resistance of the circuit R R 30 × 120 = R1 + 2 3 = 24 + = 48 W R2 + R3 30 + 120 13. (b) : Area of cross-section of the conical wire, A = geometric mean of the area of two end faces of the wire, i.e., πa2 πb2 πab × = 4 4 4 \ Resistance of conical wire, A=
R=
rl rl 4rl = = A πab / 4 πab
rl 14. (b) : As R = A So R1 =
r(4a) 2r = (2a)(a) a
15. (c) : At the centre of the square, net force on +Q is zero. But at x = 0, the charge U +Q is in unstable equilibrium. Therefore, potential energy is maximum. On either side of x-axis, potential energy –x x O decreases as shown in figure. 16. (c) : Minimum number of condensers in each row = 3000/500 = 6. If Cs is capacity of 6 condensers in a row, 1 1 1 1 1 1 = + + + + =6 Cs 1 1 1 1 1 1 or Cs = mF 6 Let there be m such rows in parallel. Total capacity = m × Cs 1 or 2 = m × \ m = 12 6 17. (c) qin 18. (b) : In Gauss’ theorem, E ∫ ⋅ dS = e0 The E on LHS of equation, i.e., the E at any point on surface is due to all the charges present in space. Here, only four point charges are given, so E is due to all four charges. 19. (b) : Point P is outside the spheres A and B and inside C. By applying principle of superposition, potential at point P, Q3 Q1 Q2 V= + + 4 πe0r 4 πe0r 4 πe0c 20. (b) : By conservation of energy q2 1 2 1 q2 q2 kr = = − 2 4 πe0 r r + r 8 πe0r ⇒
56
r(a) r = \ R1 > R2 > R3 (4a)(2a) 8a
Physics for you | November ‘15
q2
4πe0r 3
Q 21. (d) : Linear charge density of ring, λ = 2 πR q λ qQ F = 0 = 20 2 4 πe0 R 8 π e0 R
r(2a) r R2 = = (4a)(a) 2a and R3 =
k=
⇒
F Stress Y= = A Strain DR R q0Q DR = 8π2 e0 RAY
22. (b) : Since potential decreases along electric field as E0 D
E0
A
C
B
E=
25. (d) : The bridge ABCD is balanced if 10 30 = or R1 = 3 W R1 9 When this bridge is balanced, no current flows in arm BD. Therefore, R2 can have any finite value.
−dV dx
\ VA > VC and VB > VD Also due to symmetry VA = VB, VC = VD \ VA = VB > VC = VD 23. (d) :
2 I 2r I ; P = I ′2 r P1 = r = 2 4 2
P1 =
\
1 q 1 −q − 1 = 4 πe0 L 3 6 πe0 L Work done in moving charge +Q along the semicircle CRD is given by −q −qQ W = [VD − VC ](+Q) = − 0 (Q) = 6 πe0 L 6 πe0 L =
2
2
e2 r e r e = ;P = r r 4 (2R + r )2 2 R + r R + 2 2
P2 2R + r = > 1 or P2 > P1 P1 R + r
26. (b) : For potentiometer, emf E ∝ l or E = kl where k is a constant. IR Also E = ×l L E′ R \ E= × ×l (R1 + R2 + r ) L 10 5 \ E= × ×3=3V 5 + 4 +1 5 27. (d) : Resistance in parallel with voltmeter is zero. 28. (d) R − R0 , 29. (c) : Using the relation, a = T R0 DT a=
So bulb B will become brighter. R
24. (c) : +q
A
or or
–q C
B
D
2L
From figure, AC = BD = BC = L \ AD = 3L Potential at C is given by 1 q (−q ) 1 q q VC = + = − =0 4 πe0 AC BC 4 πe0 L L Potential at D is given by VD =
1 q (−q ) 1 q q + = − 4 πe0 AD BD 4 πe0 3L L
RT − R27 R27 × (T − 27)
R − R27 117 − 100 = 1000 = T − 27 = T R27 × a 100 × 1.7 × 10−4 t = 27 + 1000 = 1027°C
30. (c) : Let the resistance of bulb be R, V2 R R 2V 2 Now, RP = , \ PP = = 2P 2 R \
P=
RQ = 2R, \ PQ =
V2 P = 2R 2
3R 2V 2 2P , \ PR = = 2 3R 3 V2 RS = R, \ PS = =P R RR =
Physics for you | November ‘15
nn 57
JEE
ADVANCED
PRACTICE PAPER
2 16 CLASS
XII
Section-1
one integer Value correct Type This section contains 6 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive).
1. The given figure shows two identical parallel-plate capacitors A and B connected to a battery with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. The ratio of the electrostatic energies stored in both the capacitors before and after introduction of the dielectric is x : 10. Find the value of x. 2. An equiconvex air lens of radius of curvature 10 cm is made in an extended glass medium having refractive index m = 3/2. Find the refractive index of the material to be filled in, so that the power of the lens changes without change in its magnitude. 3. A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are mutually perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is 4 s. If one of the magnets is taken away, the period of oscillation of the other in the same field is 2x/4 s. Find the value of x. 4. A particle is uncharged and is thrown vertically upward from ground level with a speed of
5 5 m s −1. As a result, it attains a maximum height h. The particle is then given a positive charge +q and reaches the same maximum height h when thrown vertically upward with a speed of 13 m s–1. Finally, the particle is given a negative
58
Physics for you | november ‘15
charge –q. Ignoring air resistance, determine the speed (in m s–1) with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the same maximum height h ? 5. A neutron of energy 2 meV and mass 1.6×10–27 kg passes a proton at such a distance that the angular momentum of neutron relative to proton approximately equals 4×10–34 J s. The distance of closest approach neglecting the interaction between particles is given by a3×10–16 m. Find the value of a. 6. Twelve identical resistors of resistance R each are connected as shown in the figure. A Equivalent resistance between points A and E is xR . Find the value of x. 12
B F G
E
C
H D
Section-2
one or More Than one options correct Type This section contains 6 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE or MORE THAN ONE are correct.
7. Two conducting spheres of radii R and 3R carry charges Q and –2Q. Between these spheres a neutral conducting sphere of radius 2R is connected. The separation between the sphere is considerably large. Then
(a) The final charge on initially neutral conducting Q sphere is − . 3 (b) The decrease in electric potential energy of sphere of radius R is
35kQ 2 . 72R
(c) The decrease in electric potential energy of
37kQ 2 . 72R (d) The final electric potential on sphere of radius kQ . 3R will be − 6R sphere of radius R is
8. In the given circuit, R1R4 = R2R3. Then choose the correct statement(s). (a) If positions of battery and galvanometer are interchanged, then galvanometer will still show zero deflection. (b) Rate of heat dissipation through R1 will change if position of battery and galvanometer are interchanged. (c) If emf e is doubled, still no deflection is shown by galvanometer. (d) If galvanometer is not ideal, it will show deflection if emf is doubled. 9. A lens made from a material of refractive index 1.5 behaves as a converging lens in air. When placed in liquid of refractive index 8/5, it will (a) still behave as a converging lens (b) behave as a diverging lens (c) have its focal length increased (d) have its focal length decreased 10. Figure shows a square loop of side 10 cm in the x-y plane with its centre at the origin. An infinite wire is at z = 12 cm above y-axis.
Which of the following statements are correct ? (a) Resultant magnetic force on loop due to infinite wire is 6 × 10–3 N. (b) magnetic force on wire segment AB due to infinite wire is in positive x-direction. (c) magnetic force on wire segment CD due to infinite wire is in negative x-direction. (d) Torque on loop due to magnetic force is 7.2 × 10–5 N m. 11. The switches in figures (a) and (b) are closed at t = 0. Then
(a) The charge on C just after t = 0 is eC. (b) The charge on C long after t = 0 is eC. (c) The current in L just after t = 0 is e/R. (d) The current in L long after t = 0 is e/R. 12. A microammeter has a resistance of 100 Ω and a full scale range of 50 mA. It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combinations(s). (a) 50 V range with 10 kΩ resistance in series. (b) 10 V range with 200 kΩ resistance in series. (c) 5 mA range with 1 Ω resistance in parallel. (d) 10 mA range with 1 Ω resistance in parallel. Section-3
comprehension Type (only one option correct) This section contains 1 paragraph, describing theory, experiments, data etc. Two questions relate to the paragraph. Each question has only one correct answer among the four given options (a), (b), (c) and (d).
Paragraph for Question Nos. 13 and 14 Uniform electric and magnetic fields with strength E and induction B, respectively, are along y-axis as shown in figure. A particle with specific charge q/m leaves the origin O in the direction of x-axis with an initial non-relativistic velocity v0. Physics for you | november ‘15
59
13. The coordinate yn of the particle when it crosses the y-axis for the nth time is
(c)
qB2 2 π2 mn2 E
(d)
3qB2
3.
Deviation in the light ray is equal to 90°
S.
60° n=1 i
2
π mn E qB2 3π2 mn2 E qB2
14. The angle a between the particle’s velocity vector and velocity along y-axis at that moment is 3v B (a) tan −1 0 2 πnE
v B (b) tan −1 0 πnE
−1 v0 B (c) tan 2 πnE
v B (d) tan −1 0 2 πnE
This section contains 1 question, having two matching lists. Choices for the correct combination of elements from Column-I and Column-II are given as options (a), (b), (c) and (d), out of which one is correct.
15. match the columns if deviation in the column II is the magnitude of total deviation (between incident ray and finally refracted or reflected ray) to lie between 0° and 180°. Here n represents refractive index of medium. Column II 1.
(a) (b) (c) (d)
P 1,4 2,4 1 2
Q 2 1 2,4 3,4
R 3,4 1,4 2 3
S 2 3 1,4 2,4
SolutionS 1. (6) Charge on capacitor A, qA = CV 1 1 U i = CV 2 + CV 2 = CV 2 2 2 When S is opened, capacitor B will be disconnected from the battery while capacitor A will remain connected to the battery. In such a case, potential difference across A remains the same , V while the charge on B remains the same i.e., CV. Since the space between the plates of the capacitors A and B is filled with a dielectric of dielectric constant k(=3), capacitance of each capacitor is kC. 1 (CV )2 U f = UA + UB = (kC )V 2 + 2 2(kC )
Deviation in the light ray is greater than 90°
1 1 1 1 5 = CV 2 k + = CV 2 3 + = CV 2 2 k 2 3 3 Thus,
60
60°
Codes
Matching List Type
P.
4.
Charge on capacitor B, qB = CV
Section-4
Column I
60°
Speed of finally reflected or refracted light is i em erg same as speed of en tr ay incident light.
n= 2
n=2
en t ra y
(b)
R.
nc id
2 π mn E
2
Deviation in the light ray is less than 90°
zin gi
2
2.
gra
(a)
2
Q.
Physics for you | november ‘15
Ui 3 x CV 2 = = = ∴ x=6 2 5 10 U f (5 / 3)CV
2. (2) When no matter was filled, it behaves like an air lens. ∴
1 1 ma 1 2 1 1 = − 1 + = − 1 − f m g R R 3 10 10 1 2
⇒ f = –15 cm, i.e., divergent lens of focal length 15 cm. In opposite case, it will behave as convergent lens of focal length 15 cm. Let mm be the refractive index of the material to be filled. 1 mm 1 1 = − 1 − ∴ f ′ m g R R 1 2
v23
(Using eqn. (i))
= 250 – 169 = 81 ⇒ v 3 = 9 m s–1
5. (5) : L = mvlmin = 2mK lmin L ∴ lmin = 2mK 4 × 10−34 = 2 × 1.6 × 10−27 × 2 × 106 × 1.6 × 10−19 = 1.25 × 10−14 m = 125 × 10−16 m ∴ a3 = 125 or a = 5 6. (7) Points B and D have same potential, similarly F and H have same potential. The equivalent circuit is shown in figure.
1 1 2m = m − 1 + 10 10 3 ⇒
1 2 1 mv3 = 2mgh − mv22 2 2 ⇒ v32 = 2v12 − v22
1 2m m 1 = − 1 ⇒ m m = 2 5 15 3
3. (7) Time period of angular SHm of a dipole in a magnetic field, I MB
T = 2π
...(i)
Initially there are two identical dipoles at right angles to each other, so resultant magnetic moment, M0 = M 2 + M 2 = 2 M and I 0 = I + I = 2 I So, T0 = 2 π
2I ( 2 )MB
= 2π
( 2 )I MB
...(ii)
And when one magnet is taken away, the other will oscillate with period given by eqn (i), so dividing eqn (i) by (ii), T 1 = 1/ 4 T0 (2) or 4. (9)
T=
T0
1/ 4
(2)
=
4
1/ 4
(2)
= 27 / 4 s
1 2 mv = mgh 2 1 1 2 mv = mgh + qEh 2 2 1 2 mv = mgh − qEh 2 3
...(i) ...(ii)
The equivalent resistance of network is 7R/12. 7. (a, b, d) : Finally, potential on all spheres will be equal. kQ1 kQ2 kQ3 i.e., = = R 2R 3R and Q1 + Q2 + Q3 = (Q) + (−2Q) = −Q −Q −Q −Q , Q2 = , Q3 = ⇒ Q1 = 6 3 2 Change in electric potential energy of sphere of radius R = Ui – Uf k k (qi )2 − (q f )2 2R 2R 2 k 35kQ 2 2 Q = Q − = 2R 36 72R Potential on sphere of radius 3R k Q kQ = − = − 3R 2 6R =
8. (a, b, c) : On interchanging galvanometer and battery, since still R1R4 = R2R3, i.e. Wheatstone bridge condition is satisfied.
...(iii)
From equations (ii) and (iii) Physics for you | november ‘15
61
∴ No deflection in galvanometer. Initially current through R1 is e I1 = R1 + R2 Now current through R1 after interchanging e I2 = R1 + R3 Since current has changed, ∴ Rate of heat dissipation changes. Since current through galvanometer is zero, ∴ It will always show zero deflection for any value of emf. 9. (b, c) As per lens makers formula, 1 1 1 = ( a m g − 1) − fa R1 R2 fa = focal length of lens in air
L 1 1 ( m g − 1) 1 1 L = ( m g − 1) − = a fL R1 R2 ( m g − 1) f a 3/2 15 − 1 − 1 1 1 1 1 8/5 = = 16 =− . 1 fa 8 fa 3 f − 1 a 2 2
⇒ f L = −8 fa 10. (a, d) 11. (b, d) : As switch in figure (a) is closed, capacitor starts charging such that charge on C at any time t is given by q = eC (1 − e −t /CR ) .
∴ Charge on C just after t = 0 is zero and long after t = 0 (t → ∞) is ε C. As switch in figure (b) is closed, that grows a current such that current in L at any time t is e ( I= 1 − e −tR / L ) . R ∴ Current in L just after t = 0 is zero and long after e t = 0 (t → ∞) is . R 12. (b, c) : In order to increase the range of ammeter, a low resistance is connected in parallel with the ammeter. Let S be the low resistance. ∴
62
Ig Ig S = or S = G I − Ig G I − Ig
or S = (100)
50 × 10−6 5 × 10−3 − 50 × 10−6
Physics for you | november ‘15
or
S≈
100 × 50 × 10−6
5 × 10−3 Option (c) is correct.
or S = 1 Ω.
In order to change ammeter into a voltmeter, a high resistance is connected in series with the ammeter. Let the high resistance be R. ∴ I g (R + G ) = V
V 10 − G or R = − 100 Ig 50 × 10−6
or
R=
or
R = 200 k Ω − 100 Ω or R ≈ 200 k Ω
Option (b) is correct. 13. (a) : Particle’s acceleration is in y-direction, dv y qE = = constant dt m The motion of the particle is equivalent to circular motion in x–z plane with uniform acceleration in y-direction. Hence, v02 = v x2 + v z2 = constant The magnetic force cannot change the magnitude of v0 . qE The y-component of velocity, vy = t m 1 qE 2 t 2 m The time period of circular motion in x–z plane, 2πm T= qB The particle crosses y-axis after n rotations, then 2πmn t = nT = qB 2 qE 2 πmn 2 π2 mn2 E Thus, yn = × = 2m qB qB2 The y-coordinate at time t, y =
qE 2 πmn 2 πnE 14. (d): As v y = a y t = × = m qB B Thus, tan a =
v0 v B = 0 vy 2 πnE
−1 v0 B ⇒ a = tan 2 πnE
15. (a): P – 1, 4 ; Q – 2 ; R – 3, 4 ; S – 2
nn
Physics for you | november ‘15
63
Series 3 CHAPTERWISE UNIT TEST : Gravitation | Properties of Matter GENERAL INSTRUCTIONS (i) (ii) (iii) (iv) (v) (vi) (vii)
All questions are compulsory. Q. no. 1 to 5 are very short answer questions and carry 1 mark each. Q. no. 6 to 10 are short answer questions and carry 2 marks each. Q. no. 11 to 17 are also short answer questions and carry 3 marks each. Q. no. 18 is a value based question and carries 4 marks. Q. no. 19 and 20 are long answer questions and carry 5 marks each. Use log tables if necessary, use of calculators is not allowed.
1. A person sitting in a satellite of Earth feels weightlessness but a person standing on Moon has weight though Moon is also a satellite of Earth. Why? 2. A wire of length L and cross-sectional area A is made of material of Young’s modulus Y. What is the work done in stretching the wire by an amount x? 3. The velocity of water in a river is less on the bank and large in the middle; why? 4. A blackened platinum wire, when gradually heated, first appears dull red, then blue and finally white. Explain why. 5. Why iron rims are heated red hot before being put on the cart wheels? 6. When the tension in a metal wire is T1, its length is l1. When the tension is T2, its length is l2. Find the natural length of wire. 7. The excess pressure inside a soap bubble is thrice the excess pressure inside a second soap bubble. What is the ratio between the volume of the first and the second bubble? 8. A simple pendulum has a time period T1 on the earth’s surface, and T2 when taken to a height R above the earth’s surface, where R is the radius of the earth. What will be the value of T2/T1? 64
Physics for you | november ‘15
OR An astronaut on the Moon measures the acceleration due to gravity to be 1.7 m s–2. He knows that the radius of the Moon is about 0.27 times that of the Earth. What is his estimate of the ratio of the mass of the Earth to that of the Moon? (Take, acceleration due to gravity on the Earth’s surface as 9.8 m s–2) 9. A cable is replaced by another of the same length and material but of twice the diameter. (a) How does this affect its elongation under a given load? (b) How many times will be the maximum load it can now support without exceeding the elastic limit? 10. Explain why: (i) A balloon filled with helium does not rise in air indefinitely but halts after a certain height (Neglect winds). (ii) The force required by a man to raise his limbs immersed in water is smaller than the force for the same movement in air. 11. If an isotropic solid has coefficients of linear expansion, ax, ay and az for three mutually perpendicular directions in the solid, what is the coefficient of volume expansion for the solid?
12. A solid sphere of uniform density and radius R exerts a gravitational force of attraction F1 on a particle placed at P. The distance of P from the centre of the R sphere is 2R. A spherical cavity of radius is now 2 made in the sphere. The sphere (with cavity) exerts a gravitational force F2 on the same particle at P. F Calculate the ratio 1 . F2
R/2 O
P
R
14. Three capillary tubes of same radius r but of lengths l1, l2 and l3 are fitted horizontally to the bottom of a long cylinder containing a liquid at a constant head and flowing through these tubes. Find the length of the single overflow tube of the same radius r which can replace the three capillaries. 15. (a) Is it necessary that all black coloured objects should be considered black bodies? (b) Due to the change in mains voltage, the temperature of an electric bulb rises from 3000 K to 4000 K. What is the percentage rise in electric power consumed? OR (a) The line that joins the Saturn to the Sun sweeps areas A1, A2 and A3 in time intervals of 6 weeks, 3 weeks and 2 weeks respectively as shown in the figure. What is the correct relation between A1, A2 and A3? 3 weeks A2
Sun Saturn
stress F / A F ⋅l = = strain Dl / l A(Dl) The value of E depends upon nature of material. A material with higher value of E is said to be more elastic. (i) Which is more elastic, steel or rubber? Why? (ii) What values of life do you learn from this concept? 19. The distance between the centres of two stars is 10 a. The masses of these stars are M and 16 M and their radii a and 2a respectively. A body of mass m is fired straight from the surface of the larger star towards the smaller star. What should be its minimum initial speed to reach the surface of the smaller star? Obtain the expression in terms of G, M and a. OR A steel wire of length 2l is clamped between two points P and Q which are 2l apart. A body of mass M is suspended exactly from the middle of the wire. The cross-sectional area of the wire is A and its Young’s modulus is Y. In the equilibrium position, each half of the steel wire is inclined with the horizontal at an angle q. Find q. 20. A sphere is dropped under gravity through a fluid of viscosity h. Taking the average acceleration as half of the initial acceleration, show that the time to attain the terminal velocity is independent of the fluid density. OR Two rods of different materials having the same area of cross-section A are placed end to end between E=
2R
13. A box-shaped piece of gelatin dessert has a top area of 15 cm2 and a height of 3 cm. When a shearing force of 0.50 N is applied to the upper surface, the upper surface displaces 4 mm relative to the bottom surface. What are the shearing stress, shearing strain and the shear modulus for the gelatin?
6 weeks A1
(b) The time period of a satellite of Earth is 5 hours. What will be time period if the separation between the Earth and the satellite is increased to 4 times the previous value? 16. Find the velocity of efflux of water from an orifice near the bottom of a tank in which pressure is 500 gf/sq cm above atmosphere. 17. A drop of liquid of density r is floating half-immersed in a liquid of density d. If S is the surface tension, then what is the diameter of the drop of the liquid? 18. Read the passage and answer the following questions. The property of a body by virtue of which it tends to regain its original configuration as soon as the deforming forces applied on the body are removed is called elasticity. Coefficient of elasticity,
A3
2 weeks
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65
two massive walls. The first rod has a length l1, coefficient of linear expansion a1 and Young’s modulus Y1. The corresponding quantities for the second rod are l1, a2 and Y2. The temperature of both the rods is now raised by T degrees. (a) Find the force with which the rods act on each other at the higher temperature. (b) Also calculate the length of the rods at the higher temperature. solutions
1. When a person is sitting in an artificial satellite of Earth, the gravitational attraction on him due to Earth (i.e., his weight on Earth) provides the necessary centripetal force. Since the net force acting on him is zero, the person feels weightless. But when he is standing on Moon, the gravitational attraction acting on him due to Moon, is left unbalanced, which accounts for his weight on Moon. 2. Work done = elastic potential energy of stretched wire 1 = × Y × (strain)2 × volume 2 2 1 x YAx 2 = ×Y × × (A × L) = 2 L 2L 3. In the river, water flows in the form of streams. The force of adhesion is less on the streams in the middle of river than those near the bank. Due to it, the velocity of streams near the bank is least and is maximum in the middle of river.
()
l1 – l ∝ T1 If tension is T2, then l2 – l ∝ T2 l1 − l T1 = or l1T2 − lT2 = l2T1 − lT1 l2 − l T2 or
l(T1 – T2) = l2T1 – l1T2
or
l=
4S 3 × 4S = r1 r2 or r2 = 3r1 4 3 3 3 πr \ V1 = 3 1 = r1 = 1 = 1 : 27 V2 4 3 r2 3 πr 3 2 8. When pendulum is on the earth’s surface,
\ 66
F = constant or Dl ∝ F Dl If tension is T1, then Physics for you | november ‘15
4S r
\
g1 =
GM R2
When pendulum is at a height equal to R, g2 =
or
\
Excess pressure inside a soap bubble, P =
Q
5. The radius of the iron rim is smaller than the radius of the cart wheel. When the iron rim is heated, its size becomes larger than the wheel. After the rim has been planted on the wheel and is allowed to cool, it fits tightly on the wheel due to thermal contraction.
Q Y, l and A are constants,
l2T1 − l1T2 l1T2 − l2T1 = T2 − T1 T1 − T2
7. Given : P1 = 3P2
\
Fl ADl
...(ii)
Dividing eqn. (i) from eqn. (ii)
4. According to Wien’s displacement law, when blackened platinum wire is gradually heated, it first emits radiations of longer wavelengths, so it appears red. At higher temperatures, it emits blue radiations more strongly than red and appears blue. At very high temperatures, it emits all radiations strongly and appears white.
6. Y =
...(i)
GM g 1 = (2R)2 4
As T2 = 2π
Q
T2 = T1
l l and T1 = 2π g2 g1 g1 = g2
g1 =2 g1 / 4
OR g e M e Rm 2 GM e GM m ge = ; g = \ g = 2×M m Re m m Re 2 Rm 2 2 9.8 × Re 2 Me g R 2 9.8 × Re = e e2= = M m g mRm 1.7 × Rm 2 1.7 × (0.27 Re )2 9. 8 = ≈ 79 1.7 × 0.27 × 0.27
9. (a) Young’s modulus, 4 Mgl Mgl Mgl = Y= 2 = 2 πr ⋅ Dl πD 2 ⋅ Dl D ⋅ Dl π 2 where D is the diameter of the wire.
()
4 Mgl 1 i.e., Dl ∝ 2 2 πD Y D Clearly, if the diameter is doubled, the elongation will become one-fourth. πD 2 ⋅ Dl ⋅ Y i.e., Mg ∝ D 2 (b) Load, Mg = 4l Clearly, if the diameter is doubled, the wire can support 4 times the original load. Elongation, Dl =
10. (i) A balloon filled with helium goes on rising in air as long as the weight of the air displaced by it (i.e., upthrust) is greater than the weight of filled balloon. We know that the density of air decreases with height. Therefore, the balloon halts after attaining a height at which density of air is such that the weight of air displaced just equals the weight of helium filled balloon. (ii) Water exerts much more upthrust on the limbs of man than air. So the net weight of limbs in water is much less than that in air. Hence the force required by a man to raise his limbs immersed in water is smaller than the force for the same movement in air. 11. Lx = L0(1 + ax DT), Ly = L0(1 + ay DT) and Lz = L0(1 + az DT) Thus, V = LxLyLz
= L03(1 + ax DT)(1 + ay DT)(1 + az DT) Neglecting higher order terms such as axay, ayaz, azax and axayaz, we get V = V0[1 + (ax + ay + az) DT] \ Coefficient of volume expansion of the solid = ax + ay + az
12. F1 =
GMm GMm = (2R)2 4R 2
3
4 R 1 Mass removed, M ′ = π r = M 3 2 8 Force due to hollow sphere, F2 = Force due to solid sphere – Force due to removed mass M GMm G 8 m − F2 = 2 4R 2 3R 2 or
F2 =
GMm 1 4 − R 2 4 9 × 8
or
F2 =
GMm 1 1 − R 2 4 18
F2 =
GMm 9 − 2 R 2 36
7 GMm 36 R 2 F 9 F GMm 36R 2 Now, 1 = ⇒ 1= × 2 F 7 7 F2 GMm 4R 2 or
F2 =
13. A = 15 × 10–4 m2, l = 3 × 10–2 m, F = 0.50 N, Dl = 4 × 10–3 m F 0.50 Stress = = N m −2 = 333.3 N m–2 A 15 × 10 −4 −3 Dl 4 × 10 Strain = = = 0.133 − l 3 × 10 2 Shear modulus, 3 × 10 −2 stress 0.50 = = × N m −2 strain 15 × 10 −4 4 × 10 −3 1. 5 = × 105 N m −2 = 2500 N m −2 60 4
4 π Pr πPr 4 14. V1 = πPr , V2 = ...(i) and V3 = 8hl1 8h l2 8h l3 Total rate of flow, V = V1 + V2 + V3 ...(ii) If l be the length of the single overflow tube, then
πPr 4 8h l From eqns. (i), (ii) and (iii), V=
...(iii)
πPr 4 πPr 4 1 1 1 = + + 8h l 8h l1 l2 l3 or
4 Now, mass of complete sphere, M = πR 3r 3
( ) ( )
or
1 1 1 1 = + + l l1 l2 l3
or l =
l1l2l3 l1l2 + l2l3 + l3l1
15. (a) No, it is not necessary that all black coloured objects should be considered as black bodies. For example, if we take a black surface which is highly polished, it will not behave as a perfect black body. On the other hand, the sun, which is a shining hot sphere, behaves as perfect black body. (b) Electric power consumed in first case, P1 = sT14 = s(3000)4 Electric power consumed in second case, P2 = sT24 = s(4000)4 4 \ P2 = (4000) = 256 P1 (3000)4 81 Physics for you | november ‘15
67
18. (i) Steel is more elastic than rubber.
P2 − P1 P2 256 175 = −1= −1= P1 P1 81 81 \
Percentage rise in power P − P 175 = 2 1 × 100% = × 100% = 216% 81 P 1 OR (a) According to Kepler’s second law, the areal velocity of planet around the sun is constant. Therefore, A1 > A2 > A3 and A1= 2A2, A1 = 3A3. (b) Here, time period of a satellite of Earth is 5 h. By Kepler’s third law, ...(i) T2 ∝ r3 If separation is increased to 4 times, then time period, T′2 ∝ (4r)3 ...(ii) Dividing eqn. (ii) by eqn. (i) T ′ 2 (4r)3 T ′ 2 T′ = 3 , 2 = 64 or =8 2 T T r T \ T′ = 8 × T = 8 × 5 = 40 h
This is because E ∝
Under a given deforming force F, (Dl)steel < (Dl)rubber \ Esteel > Erubber (ii) In day to day life, elasticity corresponds to adjustability of the person, which depends on his nature. stress stress From, E = , we find, strain = . strain E For a given stress in life, strain will be minimum when E is maximum. It implies that a person with adjustable/flexible nature will undergo minimum strain (tension) due to stresses (ups and downs) of life. 19. Let the gravitational field due to the two stars be zero at some point O lying at a distance x from the centre of smaller star. 10a
–2
16. Pressure at orifice, P = 500 gf cm 500 = × 9.8 × (100)2 N m −2 = 500 × 98 N m −2 1000 Let h be the depth of orifice below the surface. As, P = h r g, P 500 × 98 \ h= = =5m rg 103 × 9.8 The velocity of efflux, v = 2 gh = 2 × 9.8 × 5 = 9.899 m s −1 17. As per question, Force due to surface tension + weight of liquid displaced = Weight of liquid drop 1 4 3 4 × πr dg = πr 3 rg 2 3 3 3 πr g 2πrS = [4r − 2d] 3 3 × 2πS r2 = πg[4r − 2d] 3S r2 = g (2r − d)
\ 2πrS + or or or
68
3S g (2r − d)
or
r=
\
Diameter = 2r =
12S g (2r − d)
Physics for you | november ‘15
1 (Dl)
8a
M
m O
a x
P
2a
16 M
Threshold
Then, (16 M)m Mm G 2 =G (10 a − x)2 x 1 16 = or 2 x (10 a − x)2 or 16x2 = (10a – x)2 or 4x = ± (10a – x) The negative sign is inadmissible, so x = 2a The body of mass m when fired from point P lying on the surface of heavier star must cross the threshold (the point O), otherwise it would return back. The gravitational potential energies when the body of mass m lies at positions P and O are given by 65 GMm GMm G × 16 M × m Up = − − =− 8a 2a 8a G × 16 M × m 5 GMm GMm Uo = − − =− 2a 8a 2a \ Increase in potential energy, 5 GMm 65 GMm 45 GMm DU = U p − U o = − + = 2a 8a 8a If v is the minimum speed with which the body is fired from P so as to reach O, then
45 GMm 1 2 mv = DU = 2 8a or
v=
45 GM 3 5 GM = 4 a 2 a OR
Clearly, T sin q + T sin q = Mg or 2T sin q = Mg Mg or T = ...(i) 2 sin q T Mg Stress = = ...(ii) A 2 A sin q Change in length = PCQ – PQ = 2CP – PQ =
1 − cos q 2l − 2l = 2l cos q cos q
OP l l as cos q = CP = CP or CP = cos q 2l(1 − cos q) / cos q Longitudinal strain = 2l 1 − cos q = cos q stress As, Y = longitudinal strain Mg cos q 2A sin q (1 − cos q) Mg Mg or Y = = 3 2 2A(q) (q / 2) Aq [as q is small, sin q ≈ q, cos q ≈ 1, (1 – cos q) = 2 sin2 (q/2) ≈ 2(q/2)2 ≈ q2/2] or
Y=
( )
1/3
Mg Mg q3 = or q = YA YA 20. Let r = radius of the sphere r, s = densities of the sphere and fluid respectively a = initial acceleration of the sphere when it just enters the fluid vt = terminal velocity of the sphere Net downward force (F) acting on the sphere as it just enters the fluid or
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Physics for you | november ‘15
= weight of the sphere – weight of the fluid displaced by the sphere i.e., F = 4π r 3 rg − 4π r 3 sg = 4π r 3 g(r − s) 3 3 3 4π 3 r g (r − s) (r − s)g F Thus, a = = 3 = r m 4π 3 rr 3 When the sphere attains terminal velocity, its acceleration becomes zero. Thus, a+0 a average acceleration = = 2 2 Let t be the time taken by the sphere to attain terminal velocity (vt) at From v = v0 + at, vt = 2 (as initial velocity v0 of the sphere is zero, v = vt) 2r 2(r − s)g 2 4r 2r 2v 9h or t = t = = 9h a (r − s)g / r Hence, t is independent of s (the fluid density). OR (a) When heated in free state, total expansion, Dl = Dl1 + Dl2 or Dl = a1l1T + a2l2T = (a1l1 + a2l2)T When the rod are placed end to end between the massive walls and if Dl′1 and Dl′2 are the compressions in the two rods, Fl Fl Dl1′ = 1 , Dl2′ = 2 AY1 AY2 Fl l Total compression = Dl′1 + Dl′2 = 1 + 2 A Y1 Y2 Since the total length remains unchanged, Dl′1 + Dl′2 = Dl1 + Dl2 or
F l1 l2 + = (a1l1 + a 2l2)T A Y1 Y2
or
F=
AT (a1l1 + a 2l2) (l1 / Y1 + l2 / Y2)
(b) Length of the first rod at T°C, = l1 + Dl1 – Dl′1 Fl = l1 + a1l1T − 1 AY1 l T (a l + a 2l2) = l1 + a1l1T − 1 1 1 Y1(l1 / Y1 + l2 / Y2) Similarly, length of the second rod at T° C l T (a1l1 + a 2l2) = l2 + a 2l2T − 2 Y2(l1 / Y1 + l2 / Y2)
nn
By : Prof. Rajinder Singh Randhawa*
1. Two equal masses of 6.4 kg are separated by a distance of 0.16 m. A small body of mass M = 0.10 kg is released from a point P equidistant from the two masses and a distance 0.06 m from the line joining them, as shown in figure.
(a) Calculate the velocity of this body when it passes through point Q. (b) Compute the acceleration of this body at P and Q. 2. A cosmic body P coming from infinity with velocity v0 is approaching the Sun of mass M, with its line of motion at distance d from the Sun as shown in figure. When it gets closest to the Sun, i.e. at point O, what will be its distance from the Sun?
3. A planet moves around the Sun in an elliptical orbit of semi-major axis a and eccentricity e. If the mass of the Sun is M, find the velocity of the planet at the perigee and apogee.
4. Two particles of masses m and 2m are placed at a distance d apart. Initially 2m is given a velocity u away from m and m is at rest. Find the minimum separation between the particles. 5.
The figure shows a semicircle of mass m and radius r. Find the gravitational field intensity at centre O of the semicircle ? 6. Light from a massive star suffers gravitational red shift, i.e., its wavelength changes towards the red end due to the gravitational attraction of the star. Obtain the formula for this gravitational red shift using the simple consideration that a photon of frequency u has energy hu and mass = hu/c2. 7. Two equal masses m each are hung from a balance whose scale pans differ in vertical height by h. Find the error in weighing, if any, in terms of earth’s density r. SOLUTIONS
1. (a) : According to conservation of total mechanical energy between P and Q, we have KEP + PEP = KEQ + PEQ
*Randhawa Institute of Physics, S.C.O. 208, First Fl., Sector-36D & S.C.O. 38, Second Fl., Sector-20C, Chandigarh, Ph. 09814527699 Physics for you | November ‘15
71
where M is the mass of the Sun, m is the mass of the body, r is the distance of closest approach and v is the velocity at O. Since the angular momentum of the body about the Sun will remain conserved, therefore, (mv0)d = mvr ⇒ v0d = vr ...(ii) From (i), and (ii), we get GMm 1 v0d 1 mv02 = − + m r 2 2 r
GMm GMm ⇒ 0 + − − r r
2 v 2d GM 0 Solving, we get, r = 1 + GM − 1 v02 3. Let m be the mass of the planet.
1 GMm GMm Mv 2 + − − 2 0.08 0.08 where v is the velocity of M at point Q =
Here, r = (0.06)2 + (0.08)2 = 0.1 m \
1 v2 1 = 2Gm − 2 0 . 08 0 .1
Putting the values of m and G, we get v = 6.5 × 10–5 m s–1 (b) Since the gravitational force due to two masses at point Q is zero. So, at point Q, there is no acceleration of the body. When the body is at point P, then there are two forces of equal magnitude along PA and PB. The magnitude of either force is GMm F= 2 r The resultant force along PQ is given by FR = Fcosq + Fcosq = 2Fcosq, 0.06 0.06 where, cosq = = = 0.6 r 0.1 ⇒ FR = 1.2F = =
1.2 GMm
r2 1.2 × 6.67 × 10−11 × 0.1 × 6.4
= 5.12 ×
(0.1) 10–9
Applying the conservation of angular momentum at the perigee and apogee, we get, mvprp = mvara v p ra a + c or = = ...(i) va rp a − c Using conservation of total mechanical energy, we get, GMm 1 GMm 1 mv 2p − = mva2 − rp ra 2 2 1 1 ⇒ v 2p − va2 = 2GM − rp ra r Putting v p = va a , we get, rp va =
2
(using (i) and e =
2. Applying principle of conservation of energy, KEi + PEi = KEf + PEf 1 1 GMm 0 + mv02 = mv 2 − 2 2 r Physics for you | November ‘15
...(ii)
GM a − c GM 1 − e = a a + c a 1 + e
N
Hence, acceleration = 5.12 × 10 –9 N/0.1 kg = 5.12 × 10–8 m s–2.
72
2
...(i)
Similarly, vp =
GM a
c ) a
1 + e 1 − e
4. Due to gravitational attractive force, the velocity of mass m will increase and mass 2m will decrease. Suppose at minimum separation r, velocity of both
particles is v. According to conservation of linear momentum, 2mu + 0 = mv + 2mv ⇒ v = 2u/3 ...(i) According to conservation of energy, m Gm(2m) 1 (2m) u2 − d 2 Gm(2m) 1 1 = mv 2 + (2m)v 2 − r 2 2
2m d
where u
5. Let mass of given semicircle be m. Consider a small element of length rdq as shown in figure. The mass of this element is y
d O r
c2
is the mass of the photon
dI =
Gm πr 2
r2
...(i)
dq (using (i))
dq
By symmetry, the net field along y-axis is zero. The component along negative x-axis is dIsinq The resultant field intensity is given by π Gm
I = ∫ dI sin q = ∫
0
I=
Gm πr
2
πr
2
sin q dq =
[− cos π + cos 0] =
^ \ I = − (2Gm / πr 2 ) i
2Gm πr 2
Gm πr
2
c c GM = 1− λ ′ λ Rc 2
7. Since acceleration due to gravity varies near the earth surface as 2h g (h) = g 1 − R m
Gravitational field intensity at centre O due to this Gdm
u or
GM λGM λ ′ = λ 1 + 2 or | λ ′ −λ |= Rc Rc 2
x
m m (rdq) = dq πr π
element is dI =
Rc
2
GM or λ ′ = λ 1 − 2 Rc GM If < < 1, then using binomial theorem Rc 2
h1
dm =
GM
−1
2
(rd)
hu
GM hu , R c 2
or u′ = u −
2Gm2 3 2 2Gm2 mu − = m u − d r (using (i)) 2 3 1 ⇒ r= 1 u2 − d 6Gm 2
hu′ = hu −
[− cos q]0π
along negative x -axis.
6. If hu is the energy of the photon neglecting the effect of gravitational attraction and hu′ is the energy after the red shift, then
h
m
h2
From figure h1 > h2, so W1 will be lesser than W2 h h and W2 – W1 = mg2 – mg1 = 2 mg 1 − 2 R R GM h or W2 − W1 = 2m R2 R GM As g = 2 and (h1 − h2 ) = h R 2mhG 4 3 or W2 − W1 = × πR r 3 R3 4 3 M = r × πR 3 8 \ W2 − W1 = πrGhm 3
nn
Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time -Thomas A. Edison
Physics for you | November ‘15
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Series 6 CHAPTERWISE PRACTICE PAPER : Semiconductor Electronics - materials devices and simple circuits | Communication Systems Time Allowed : 3 hours
Maximum Marks : 70 GENERAL INSTRUCTIONS
(i) All questions are compulsory. There are 26 questions in all. (ii) This question paper has five sections: Section A, Section B, Section C, Section D and Section E. (iii) Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each. (iv) There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.
section-A 1. Define the term critical frequency in relation to sky wave propagation of electromagnetic waves. 2. Name two factors on which electrical conductivity of a pure semiconductor at a given temperature depends. 3. Give one example each of a system that uses the (i) Skywave (ii) Space wave mode of propagation. 4. What happens to the width of depleting layer of a p-n junction when it is (a) forward biased (b) reverse biased? 5. If the base region of a transistor is made large as compared to a usual transistor, how does it affect (a) the collector current (b) current gain of this transistor ? section-b
6. Why is communication using line of sight mode limited to frequencies above 40 MHz? 7. A semiconductor is known to have an electron concentration of 8 × 1013 per cm3 and a hole concentration of 5 × 1012 per cm3. 74
Physics for you | November ‘15
(i) Is the semiconductor n-type or p-type? (ii) What is the resistivity of the sample if the electron mobility is 23,000 cm2 V–1 s–1 and hole mobility is 100 cm2V–1s–1? 8. A diode detector, with a load resistance R = 250 kW in parallel with a capacitor C = 100 pF, is used to detect an amplitude modulated carrier. Find the highest modulation frequency that can be detected without excessive distortion. 9. The output of an AND gate is connected to both the inputs of a NAND gate. Draw the logic circuit of this combination of gates and write its truth table. 10. In half-wave rectification, what is the output frequency if the input frequency is 50 Hz ? What is the output frequency of a full-wave rectifier for the same input frequency? OR A Zener of power rating 1 W is to be used as a voltage regulator. If Zener has a breakdown of 5 V and it has to regulate voltage which fluctuates between 3 V and 7 V, what should
be the value of Rs for safe operation as per the given figure?
section-c
11. What is a carrier wave? Why high frequency carrier waves are employed for transmission? 12. From the output characteristics of common emitter circuit shown in figure, calculate the values of bac and bdc of the transistor when VCE is 10 V and IC = 4.0 mA.
13. Draw a block diagram for a simple amplitude modulator and explain briefly how amplitude is modulation achieved. 14. State the factor, which controls (i) wavelength of light and (ii) intensity of light, emitted by a LED. Write two advantages of LED over incandescent lamp. 15. Write the symbol and truth table of an AND gate. Explain how this gate is realised in practices by using two diodes. 16. Consider the circuit arrangement shown in figure (i) for studying input and output characteristics of n-p-n transistor in CE configuration.
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Select the values of RB and RC for a transistor whose VBE = 0.7 V, so that the transistor is operating at point Q as shown in the characteristics shown in figure (ii). Given that the input impedance of the transistor is very small and VCC = VBB = 16 V, also find the voltage gain and power gain of circuit making appropriate assumptions. OR If each diode in given figure has a forward bias resistance of 25 W and infinite resistance in reverse bias, what will be the values of the current I1, I2, I3 and I4? A C E I1 G
I4
125
I3
125
I2
125 25
5V
B D F H
17. In the circuit shown in figure, when the input voltage of the base resistance is 10 V, VBE is zero and VCE is also zero. Find the values of IB, IC and b. 18. Suppose a n-type wafer is created by doping Si crystal having 5 × 1028 atoms per m3 with 1ppm concentration of As. On the surface 200 ppm concentration of Boron is added to create p region in this wafer. Considering ni = 1.5 × 1016 m–3, (i) Calculate the densities of the charge carriers in the n and p regions. (ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased. 19. Write the truth table for the circuits given in figure consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.
20. An amplitude modulated wave is as shown in figure. Calculate (i) the percentage modulation, (ii) peak carrier voltage and, (iii) peak value of information voltage. V
20 V 100 V t
21. An n-p-n transistor is connected in CE configuration in which collector supply is 8 V and the voltage-drop across the load resistor of 800 W connected in the collector circuit is 0.8 V. If the current amplification factor is 25, determine collector-emitter voltage and base current. If the internal resistance of the transistor is 200 W, calculate the voltage-gain and power-gain. 22. An amplitude modulated wave is represented as cm(t) = 5(1 + 0.6 cos 6280 t) sin 211 × 104 t, volts. (i) What are the minimum and maximum amplitudes of the A.M. wave? (ii) What frequency components are contained in the modulated wave? (iii) What are the amplitudes of the components? section-D 23. Two students namely Shobit and Amit were asked to take up a project on efficient lighting for road ways, cycle paths and bus lanes. They found LED is the best source for the above said reasons. They collected the information from various sources and submitted the project about its working, advantages and its applications by presenting with a good working model. (a) By seeing these two students, what kind of qualities do you want to adopt from them? (b) Explain LED with neat diagram and draw its symbol. section-e
24. Explain briefly, with the help of circuit diagram, how V-I characteristics of a p-n junction diode are obtained in (i) forward bias, and (ii) reverse bias. Draw the shape of the curves obtained. OR Discuss common emitter amplifier, using n-p-n transistor. Find its current gain, voltage gain and power gain.
25. Draw the energy band diagrams of p-type and n-type semiconductors. Explain with a circuit diagram the working of full-wave rectifier. OR What are energy bands? How are these formed? Distinguish between a conductor, an insulator and a semiconductor on the basis of energy band diagram. 26. Draw a labelled circuit diagram of a common emitter amplifier using a p-n-p transistor. Explain how the input and output voltages are out of phase by 180° for a common-emitter transistor amplifier. Define the term voltage gain and write an expression for it. OR Explain briefly, with the help of a circuit diagram, how a p-n junction works as a half wave rectifier. Explain with the help of a circuit diagram how a Zener diode works as a dc voltage regulator. Draw its I-V characteristics. solutions 1. Critical frequency is the highest frequency of the radiowaves which when sent normally towards the given layer of ionosphere gets reflected from ionosphere and returns to the earth. It is given by uc = 9 (Nmax)1/2 where Nmax is the maximum number density of electrons in the given layer of ionosphere. 2. (a) Width of the forbidden gap (Eg). (b) Intrinsic charge carrier concentration (ni). 3. (i) Short wave broadcast (ii) Television transmission. 4. (a) Decreases (b) Increases.
5. When the base region becomes large, most of the charge carriers coming from emitter will be neutralized in the base by the electron-hole combination. Due to this (a) collector current decreases and as a result of this (b) current gain also reduces. 6. At frequencies above 40 MHz, communication is limited to line-of-sight communication. At these frequencies, the sizes of transmitting and receiving antennas are relatively smaller and can be placed at heights of many wavelengths above the ground. During line of sight communication, the waves coming directly from transmitting antenna towards receiving antenna get blocked at some point by the curvature of the earth. Physics for you | November ‘15
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7. (i) Given ne = 8 ×1013 cm–3, nh = 5 × 1012 cm–3 Since ne > nh, the semiconductor is n-type. (ii) As me = 23,000 cm2 V –1s–1, mh = 100 cm2 V –1s–1 s = e(neme + nhmh)
= (1.6 × 10–19)[(8 × 1013) (23,000) + (5 × 1012) × 100] –1 –1 = 0.29448 W cm 1 1 Thus, ρ = = = 3.396 W cm s 0.29448 W−1 cm −1
8. Here, R = 250 kW = 250 × 103 W = 2.5 × 105 W, C = 100 pF = 100 × 10–12 F = 1 × 10–10 F 1 For satisfactory detection, > RC uc or uc <
1 1 = = 40 kHz 5 RC (2.5 × 10 W)(10−10 F)
Thus, the highest modulation frequency that can be detected with permissible distortion is 40 kHz. 9. The logic circuit for the given combination of gates is shown in the figure. Y = A.B
A B
AND Gate
Y
OR Given P = 1 W, Vz = 5 V Vi(max) = 7 V P 1 Iz = = = 0.2 A Vz 5 For safe operation, Vi(max) − Vz 7 − 5 2 Rs = = = = 10 W Iz 0.2 0.2 11. A carrier wave is an electromagnetic wave of high frequency and of constant amplitude, which is employed to carry the audio signals from one location to other on the surface of earth. For the transmission of audio signals, the high frequency carrier waves are used, because the effective power radiated by a longer wavelength 2
NAND Gate
Truth table A
B
Y′ = A · B
Y = Y ′ .Y ′
0 0 1 1
0 1 0 1
0 0 0 1
1 1 1 0
l base-band signal should be small as P ∝ . λ Thus, for a good transmission, we need high powers and hence we need high frequency carrier waves to carry the base-band signal (message signal). 12. Consider any two characteristics for two values of IB which lie above and below the given value of IC. Here IC = 4.0 mA, therefore we select the two characteristics for IB = 20 mA and 30 mA.
10. In half wave rectification, only one ripple is obtained per cycle in the output.
Output frequency of a half wave rectifier = input frequency = 50 Hz In full wave rectification, two ripples are obtained per cycle in the output. Output frequency = 2 × input frequency = 2 × 50 = 100 Hz 78
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From the graph, at VCE = 10 V, DIB = (30 – 20) mA = 10 mA DIC = (4.5 – 3.0) mA = 1.5 mA DIC 1.5 mA = = 150 Therefore, bac = DIB 10 mΑ At VCE = 10 V, either (i) IB = 30 mA and IC = 4.5 mA or (ii) IB = 20 mA and IC = 3 mA.
Therefore 4.5 mA I = 150 For (i), bdc = C = IB 30 mA 3 mA IC = = 150 For (ii), bdc = IB 20 mΑ 13. The block diagram is shown here.
(ii) Truth table
Bandpass AM x(t) Square y(t) m(t) wave filter + law device Am sin mt centred at c c(t) (Modulating Ac sin ct 2(t) Bx ( t ) + Cx signal) (Carrier wave)
The modulating signal is superposed on carrier wave of high frequency. Let the signal produced be x(t) = Amsinwmt + Acsinwct ...(i) The resultant wave so obtained is sent to square law device which produces wave y(t) = Bx(t) + Cx2(t) ...(ii) where B and C are arbitrary constants. From equations (i) and (ii), we get C 2 y(t) = BAmsinwmt + BAc sinwct + (Am+ A2c ) 2 C C 2 − Am cos 2wmt − Ac2 cos 2wct 2 2 + CAmAccos(wc – wm)t – CAmAccos(wc + wm)t C 2 ( A + Ac2 ) and 2 m sinusoidal waves of frequencies wm, 2wm, wc, 2wc , (wc – wm) and (wc + wm). This is finally sent to bandpass filter which rejects dc and sinusoids of frequencies wm , 2wm and 2wc allows the frequencies wc, wc – wm and wc + wm. The output of bandpass filter is an amplitude modulated wave. In this equation, there is dc term
14. (i) Wavelength of light emitted depends on the nature of semiconductor. (ii) Intensity of light emitted depends on the forward current. Advantages of LED over incandescent lamp are as follows (a) low operational voltage and less power. (b) fast action and no warm-up time required.
B
B 0 1 0
Y=A.B 0 0 0
1
1
1
(iii) Realisation of AND gate
+5V +5V
(a) When A = 0, B = 0 Both the diodes D1 and D2 conduct. Potential at Y is zero as most of potential falls across the resistance R. (b) When A = 1, B = 0. Diode D1 do not conduct but D2 conducts. Voltage of 5 V drop across R. Now potential at Y is 0. (c) When A = 0, B = 1. D1 conducts, D2 do not conduct. Voltage drop of 5 V across R. Potential at Y is 0. (d) When A = 1, B = 1 D1 and D2 both do not conduct, no current flow through R. Hence output Y = 1 16. From the output characteristics at point Q, VCE = 8 V and IC = 4 mA VCC = ICRC + VCE 16 − 8 V − VCE , RC = = 2 kW RC = CC IC 4 × 10−3 Since, VBB = IBRB + VBE 16 − 0.7 RB = = 510 kW 30 × 10−6 I 4 × 10−3 = 133 Now, b = C = I B 30 × 10−6 R 2 × 103 Voltage gain, AV = b C = 133 × = 0.52 RB 510 × 103 R Power gain, AP = b × AV = b2 C RB
15. (i) The symbol of AND gate is as follows. A
A 0 0 1
Y
= (133)2 ×
2 × 103
510 × 103
= 69
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OR I3 is zero as the diode in that branch is reverse biased. Resistance in the branch AB and EF are each (25 + 125) W = 150 W. A C E I1 G
17.
I4
125
I3
125
I2
125 25
5V
B D F H
As AB and EF are identical parallel branches, their 150 = 75 W effective resistance is 2 \ Net resistance in the circuit = (75 + 25) W = 100 W. 5 = 0.05 A \ Current I1 = 100 As resistances of AB and EF are equal, 0.05 \ I2 = I4 = = 0.025 A 2
Given Vi = 10 V, RB = 400 kW = 400 × 103 W RC = 3 kW = 3 × 103 W, VBE = 0 VCE = 0, VCC = 10 V As Vi – VBE = RBIB \ 10 – 0 = (400 × 103)IB 10 IB = = 25 × 10−6 A = 25 mA 3 400 × 10 and VCC – VCE = ICRC 10 IC = = 3.33 × 10−3 = 3.33 mA 3 × 103 I 3.33 × 10−3 b= C = = 133 IB 25 × 10−6 18. (i) For n-type region, 1 ne = ND = × 5 × 1028 = 5 × 1022 m–3 6 10 1 1 ppm = 106 As nenh = ni2, 80
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n 2 (1.5 × 1016 m−3 )2 nh = i = = 0.45 × 1010 m–3 ne 5 × 1022 m−3 For p-type region, 200 25 –3 nh = NA = × 5 × 1028 = 1 × 10 m 6 10 ni2 (1.5 × 1016 m −3 )2 = Now, ne = = 2.25 × 107 m–3 nh 1 × 1025 m −3
(ii) The minority carrier holes of n-region wafer (nh = 0.45 × 1010 m–3) would contribute more to reverse saturation current than minority carrier electrons of p–region wafer (ne = 2.25 × 107 m–3) when p – n junction is reverse biased. 19. Boolean expression for logic circuit (a) Y=A
A
Here the given NOR gate with A short circuit input is acting as 0 1 NOT gate. Its truth table is Boolean expression for logic circuit (b)
Y
1 0
So by De-Morgan’s theorem
(
)
Y = A + B = A⋅B = A⋅B Hence, the logic circuit acts like AND gate. Its truth table is A 0 0 1 1
B 0 1 0 1
Y 0 0 0 1
20. According to diagram, Vmax = V
100 20 = 50 V , Vmin = = 10 V 2 2 20 V 100 V
t
(i)
Percentage modulation, P = m × 100 − Vmin V × 100 P = max Vmax + Vmin
where m = modulation index 50 − 10 2 = × 100 = × 100 = 66.7% 50 + 10 3 (ii) Peak carrier voltage, Vc = Vmax + Vmin 2 50 + 10 = = 30 V. 2 (iii) Peak information voltage = Vm = mVc 2 = × 30 = 20 V. 3 21. Here, VCC = 8 V, RC = 800 W, ICRC = 0.8 V, b = 25, ri = 200 W 0.8 V = 1.0 × 10−3 A As ICRC = 0.8 V, IC = 800 W Further, VCE = VCC – ICRC = 8 V – 0.8 V = 7.2 V I I 1.0 × 10−3 Since b = C , I B = C = A IB b 25
= 0.04 × 10–3 A = 40 mA 800 W R Voltage gain, AV = b out = 25 = 100 Rin 200 W (As Rout = RC = 800 W and Rin = ri = 200 W) Power gain, AP = bAV = 25(100) = 2500 22. Given the A.M. wave, cm(t) = 5(1 + 0.6 cos 6280 t) sin 211 × 104 t, volts. Comparing with the standard A.M. wave, cm(t) = Ac(1 + m cos wm t) sin wc t, we get Ac = 5 V, m = 0.6 w 6280 Modulating frequency, fm = m = = 1 kHz 2π 2π
w 211 × 104 Carrier frequency, fc = c = = 336 kHz 2π 2π (i) Minimum amplitude of A.M. wave = Ac – mAc = 5 – 0.6 × 5 = 2 V. Maximum amplitude of A.M. wave = Ac + mAc = 5 + 0.6 × 5 = 8 V. (ii) Frequency components of the A.M. wave are fc – fm, fc, fc + fm i.e., 336 – 1,336,336 + 1 or 335 kHz, 336 kHz, 337 kHz. (iii) The amplitudes of the three components are mAc mA 0. 6 × 5 0. 6 × 5 , Ac , c i.e., , 5, 2 2 2 2 or 1.5 V, 5 V, 1.5 V.
23. (a) Initiative, curiosity, scientific awareness, community service. (b) Light emitting diode (LED) is a junction diode made of gallium arsenide or indium phosphide in which when holeelectron pairs recombine at forward biased p-n junction, energy is released in the form of light. The principle on which LED works is spontaneous emission of radiation, when an electron jumps from higher energy level to lower energy level in a semiconductor atom. At forward biased p-n junction, free electrons of n-type combine with holes of p-type semiconductor. As free electrons lie in conduction band and holes lie in valence band, so electron falls from the higher to lower energy level containing holes and the energy is released in the form of radiation. The energy of radiation emitted by LED is equal to or less than the forbidden energy band gap Eg of the semiconductor used, and is given hc by hu = = Eg . λ The frequency of emitted radiation in LED, thus depends upon the band gap energy Eg of the semiconductor used. The intensity of emitted radiation in LED depends upon the forward current flowing through the LED. More the forward current flowing through the LED, more the hole-electron pairs combine at forward biased p-n junction, releasing more number of photons and hence larger will be the intensity of emitted radiation. 24. Refer point 9.3(4,5) page no. 588 (MTG Excel in Physics). OR
Refer point 9.4(8) page no. 595 (MTG Excel in Physics). 25. Refer point 9.2, (2 (a, b)) page no. 585 and point 9.3(6(ii)) page no. 588 (MTG Excel in Physics) OR Refer point 9.1 (3,4,5,6) page no. 583 (MTG Excel in Physics) 26. Refer point 9.4 (9) page no. 595 (MTG Excel in Physics) OR Refer point 9.3(6(i), 7(i)) page no. 589 (MTG Excel in Physics) nn Physics for you | November ‘15
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Y U ASK
WE ANSWER Do you have a question that you just can’t get answered? Use the vast expertise of our mtg team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough. The best questions and their solutions will be printed in this column each month.
Q1. When you load up a plastic shopping bag with groceries and then carry the bag by the loops at the top of the bag, why will the loops initially withstand the load but then, several minutes later, begin to stretch, perhaps to the point of tearing? – Akash Jain (UP)
Ans. If you suspended a load from the lower end of a spring hanging from a ceiling, the spring will stretch by a certain amount and then stay stretched. Plastic, which consists of polymers, is different. If you suspend a load from the lower end of a plastic strip, the strip will initially stretch like the spring but thereafter it will gradually stretch more in what is called viscoelastic creep. The mechanism of this creep can vary from polymer to polymer. The polymer consists of many long and entangled molecules. When the polymer is put under load, these molecules gradually disentangle somewhat because very are pulled in the direction of the load. The gradual reorientation of the molecules allows the plastic to gradually stretch. Q2. What causes the tides? Why do most shore locations have two high tides per day but others have only one? – Sahil Verma (Delhi)
Ans. The primary cause of tides is the Moon’s gravitational pull on Earth’s oceans even though that force is not strong enough to lift the water. Because the force varies over Earth’s surface (strongest on the side facing the Moon, weakest on the opposite side), the force reshapes the water distribution by stretching it parallel to the line connecting Earth and the Moon. The stretching 82
Physics for you | November ‘15
produces two bulges in the water distribution, one on the side facing the Moon and one on the opposite side. If Earth did not rotate, then a shore location in the bulge facing the Moon would have high water (high tide) all day and a shore location in the opposite bulge would as well. However, Earth’s rotation means that a shore location rotates through both bulges in about a day and will thus have two high-water intervals. The bulges are not exactly positional on a line through Earth and the Moon because the water motion encounters friction within the water and against shorelines. The friction delays the water’s response to the stretching by the Moon. So, the high-water point in a port city may occur an hour or more after the Moon is highest in the sky. Another complicating factor is that the gravitational force from the Sun also tends to stretch the water distribution. However, the solar effect is, roughly less than half the lunar effect. Although the Sun is much larger than the Moon, it is also much farther from Earth. During New Moon and Full Moon, the Sun and Moon are aligned and their tidal effects sum to give larger tides called spring tides. When the directions to the Sun and Moon are separated by 90°, the sum gives neap tides. Because of these various complications some shore locations can have only one noticeable tide per day. Q3. When an airplane flies somewhat overhead and close enough to be heard, lower your head by stooping to the ground. Why does the frequency of the airplane noise increases as you lower your head ? – Shreyas (Haryana)
Ans. The sound you hear consists of the sound coming directly to you from the airplane and the sound that reflects to you from the ground. The two sets of sound waves undergo interference at your ears and you hear primarily the waves that constructively interfere, they reinforce one another rather than cancel one another. The height above the ground at which constructive interference occurs depends on the wavelength, greater wavelength requires greater height. As you lower your head, you move down to the heights at which shorter wavelengths or higher frequencies, undergoes constructive interference. So, as you stoop, the sound you hear increases in frequency.
to d jo in tl y s a w a rd e of ry ve 2015 wa e disco nald for th s. P ri z o s l D a e c m b M o e . N v B he Arthur trinos ha d u n e a n a t s a jit a te th e laurea shows Takaaki K utrinos, th s, which e n n o f ti o a ld u ill re o c tu c s o that ike na neutrino ameleon-l le physics ring the ch le in partic zz ademy of u c p A g h in is By uncove l Swed ng-stand ya lo o R a e d e th lv , had so e cosmos ch rasp of th ctions, su alter our g uclear rea n a in d d . e id te in a a a s o rem les cre Sciences cule partic the neutrin s is s a e in w d m a it c t re e a a d th For d Neutrinos hers prove the stars. s sun and an researc c e laureate ri e th m d A n l a as in the ti eutrinos e n article un th p f o l g a n s c lli d ti e e in k r disp hypoth are three to anothe 56. There one kind m o real in 19 fr . s te s a massle ey oscill inos were showed th that neutr n o ti o n long-held
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NASA plans to set up Mars colonies by 2030
Live Physics
ize 2015 Nobel Pr e in P h y s ic s
N
ISRO will launch India’s own satellite navigation: IRNSS
T
he Global Positioning System used by almost all countries in the world is controlled by USA Govt. American monopoly on satellite based navigation is all set to end now, as Indian Space Research Organisation will launch India’s own satellite based navigation next year. Christened as Indian Regional Navigation Satellite System or IRNSS, this navigation system would be controlled by Indian Govt. Scientists at ISRO started a series of meetings with various location and navigation device manufacturers, mobile phone companies and global information system (GIS) technology developers from all over the world, to explain them the advantages and benefits of using IRNSS. As per reports coming in, ISRO will offer two types of services via IRNSS; Standard Positioning Service (SPS), which would be available for all users, mobile phones, e-commerce services, digital services and Restricted Service (RS),which would be exclusively for defense users.
ASA is planning to have humans living on Mars in the next few decades. Moving to have Earth independent colonies on the Red Planet will be the end point of years of research, the agency has said, but it plans for that to be complete by the 2030s. NASA laid out the plans in a large document: “NASA’s Journey to Mars -Pioneering Next Steps in Space Exploration”. The document lays out the three stages of NASA’s plan to get to Mars. The first is Earth Reliant; the second is Proving Ground, where the operations will be tested out in deep space, but in an environment that allows humans to get back to Earth in days. NASA hopes that those two first stages allow it to get to the Earth independent stage. It sees the Earth independent colonies as being a global achievement that marks a transition in humanity’s expansion as we go to Mars not just to visit, but to stay. While there, humans will live and work within habitats that support human life for years, with only routine maintenance. They’ll harvest resources to create fuel, water, oxygen and building materials and use advanced communication systems to send information back with only a 20-minute delay . NASA has expressed such interest before, most recently proposing to send a small greenhouse to the planet in order to experiment with cultivating plant life, something that would be essential to establishing a permanent colony in the future. Although Mars does not currently seem to be a great habitat for existing life. It is still possible, things may be living beneath the surface something that can only be explored effectively by humans, not robots.
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across
1. A region under the influence of some physical agency. (5) 3. An abnormal transient electrical disturbance in a conductor. (5) 5. Rotating part of electric motor or generator. (8) 7. A penlike device attached to a VDU by which information may be input to a computer. (5, 3) 8. A manually operated device by which people communicate with computer. (8) 9. A mechanical device that prevents any sudden or oscillatory motion of a moving part of any piece of apparatus. (7) 13. A device that allows microwave radiation to pass in one direction, while absorbing it in the reverse direction. (8) 15. The converse of compression. (11) 17. A type of circuit having two stable states. (8) 19. A thread like body, particularly the conductor of metal or carbon in an incandescent lamp. (8) 22. X-rays of long wavelength produced when electrons are accelerated by voltages of 25 kV or less. (5, 4) 24. A conductor or group of conductors used for the transmission and/or distribution of electrical power. (4) 25. A device incorporated in an electrical or other indicating instrument to provide the necessary damping. (6)
down
1. A system of wires or waveguides that conveys radio frequency power between a radio aerial and a transmitter or reciever, with minimum loss. (6) 2. A planoconcave lens placed between the objective and eyepiece in a telescope to increase magnification. (6, 4) 4. Process of boiling. (10) 6. The adjustment of tuning of the notes of a keyboard instrument to give a near diatonic scale for all keys. (11)
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10. The science concerned with the production, properties and propagation of sound wave. (9) 11. A device used for stabilizing voltage, consisting of a sensitive metallic resistor whose resistance increases with temperature. (9) 12. To bring neutrons into thermal equilibrium with surrounding. (10) 14. The number of hydrogen (or equivalent) atoms that an atom will combine with or displace. (7) 16. The study of the production and effects of very low temperatures. (10) 18. A unit of work or energy equal to one watt operating for one hour. (4, 4) 20. An acronym for allied submarine detection investigation committee. (5) 21. The disappearance of signals for short time due to variations in the height and density of ionisation of ionosphere. (6) 23. A missile or space vehicle powered by ejecting gas. (6)
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